Linear Programming

The given data can be shown in a table as follows:

Now, let the required weekly production of gadgets A and B be x and y respectively

As it is given that profit on each gadget A is Rs. 30 and that on B is Rs. 20, so profit on x and y number of gadgets A and B respectively are 30x and 20y.

If z = total profit then we have,

z = 30x + 20y

Further, it is also given that the production of A and B requires 10 hours per week and 6 hours per week in the foundry. Also, the maximum capacity of the foundry is given as 1000 hours.

Now, x units of A and y units of B will require 10x + 6y hours

So, we have

10x + 6y ≤ 1000

This is our first constraint.

Given, production of one unit gadget A requires 5x hours per week and y units of gadget B requires 4y hours per week, but the maximum capacity of the machine shop is 600 hours per week.

So, 5x + 4y ≤ 600

This is our second constraint.

Hence, the mathematical formulation of LPP is:

Find x and y which will maximize z = 30x + 20y

Subject to constraints,

10x + 6y ≤ 1000

5x + 4y ≤ 600

and also, as production cannot be less than zero, so x, y ≥ 0

The given data can be shown in a table as follows:

Let production of product A be x units and of be B y units.

Given,

Profit on 1 unit of product A = Rs.60

Profit on 1 unit of product B = Rs.80

So, profit on x units of A and y units of B is 60x and 80y respectively.

Let z = total profit,

So, we have

z = 60x + 80y

Given, a minimum supply of product B is 200

So, y ≥ 200 (First constraint)

Given that, production of one unit of product A requires 1 hour of machine hours, so x units of product A require x hours but total machine time available for product A is 400 hours

So, x ≤ 400 (Second constraint)

Given, each unit of product A and B requires one hour of labour hour, so x units of product A require x hours and y units of product B require y hours of labour hours, but total labour hours available is 500 so

x + y ≤ 500 (Third constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

Minimize z = 60x + 80y

Subject to constraints,

y ≥ 200

x ≤ 400

x + y ≤ 500

and also, as production cannot be less than zero, so x, y ≥ 0

The given data can be formulated in a table as below.

Let, required production of product A, B and C be x, y and z units respectively.

Given, profit on one unit of product A, B and C are Rs 3, and Rs 2, Rs 4.

So, profit on x, y, z units of A, B, C Rs 3x, Rs 2y, Rs 4z.

Let U be the total profit, so

U = 3x + 2y + 4z

Given, one unit of product A, B and C requires 4, 3 and 5 minutes on machine M₁. So, x units of A, y units of B and z units of C need 4x, 3y and 5z minutes. Maximum capacity on machine M₁ is 2000 minutes, so,

4x + 3y + 5z ≤ 200 0 (First constraint)

Given, one unit of product A, B and C requires 2, 2 and 4 minutes on machine M₂. So, x units of A, y units of B and z units of C require 2x, 2y and 4z minutes. Maximum capacity on machine M₂ is 2500 minutes, so,

2x + 2y + 4z ≤ 250 0 (Second constraint)

Also, given that firm must manufacture more than 100 A’s, 200 B’s, 50 C’s also not more than 150 A’s, so,

100 ≤ x ≤ 150,

y ≥ 200 (Other constraints)

z ≥ 50

Hence, mathematical formulation of LPP is:

Find x, y and z which maximize U = 3x + 2y + 4z

Subject of constraints,

4x + 3y + 5z ≤ 2000

2x + 2y + 4z ≤ 2500

100 ≤ x ≤ 150,

y ≥ 200

z ≥ 50

and also, as production cannot be less than zero, so x, y ≥ 0

Given equation can be written in tabular form as

Let required production of product A be x units and product B be y units.

Given, profit on one unit of product A and B are Rs 2 and Rs 3 respectively, so profits on x units of product A and y units of product B will be Rs 2x and Rs 3y respectively.

Let total profit be Z, so Z = 2x + 3y

Given, production of one unit of product A and B require 1 and 1 minute on machine M₁ respectively, so production of x units of product A and y units of product B require x minutes and y minutes on machine M₁ but total time available on machine M₁ is 600 minutes, so

x + y ≤ 400 (First constraint)

Given, production of one unit of product A and B require 2 minutes and 1 minutes on machine M₂ respectively. So, production of x units of product A and y units of product B require 2x minutes and y minutes respectively on machine M_2, but machine M₂ is available for 600 minutes, so

2x + y ≤ 600 (Second constraint)

Hence, the mathematical formulation of LPP is:

Find x and y which

maximize Z = 2x + 3y

Subject to constraints,

x + y ≤ 400

2x + y ≤ 600

and, x, y ≥ 0 [Since production of the product cannot be less than zero]

Let plant 1 requires x days, and plant II requires y days per month to minimize cost.

Given, plant I and II costs Rs 2500 per day and Rs 3500 per day respectively, so cost to run plant I and II are Rs 2500x and Rs 3500y per month.

Let Z be the total cost per month,

So, Z = 2500x + 3500y

Given, production of tyre A from plant I and II is 50 and 60 respectively, so production of tyre A from plant I and II will be 50x and 60y respectively per month but the maximum demand of tyre A is 2500 per month so,

100x + 60y ≥ 2500 [First constraint]

Given, production of tyre B from plant I and II is 100 and 60 respectively, so production of tyre B from plant I and II will be 100x and 60y per month respectively but the maximum demand of tyre B is 3000 per month, so

100x + 200y ≥ 3000 [Second constraint]

Given, production of tyre C from plant I and II is 100 and 200 respectively. So, production of tyre C from plant I and II will be 100x and 200y per month respectively but the maximum demand of tyre C is 7000 per day, so

100x + 200y ≥ 7000 [Third constraint]

Hence, mathematical formulation of LPP is,

Find x and y which

Minimize Z = 2500x + 3500y

Subject to constraint,

50x + 60y ≥ 2500

100x + 60y ≥ 3000

100x + 200y ≥ 7000

And, x, y ≥ 0 [Since number of days cannot be less than zero]

Let required production of product A be x units and production of product B be y units.

Given, profits on one unit of product A and B are Rs 60 and Rs 40 respectively, so profits on x units of product A and y units of product B are Rs 60x and Rs 40y.

Let Z be the total profit, so
Z = 60x + 40y

Given, production of one unit of product A and B require 5 hours and 3 hours respectively man hours, so x unit of product A and y units of product B require 5x hours and 3y hours of man hours respectively but total man hours available are 45000 hours, so

5x + 3y ≤ 45000 (First constraint)

Given, demand for product A is maximum 7000, so

x ≤ 7000 (Second constraint)

Hence, mathematical formulation of LPP:

Find x and y which

maximize Z = 60x + 40y

Subject to constraints,

5x + 3y ≤ 45000

x ≤ 7000

y ≤ 10000

x, y ≤ 0 [Since production cannot be less than zero]

Let x and y be the packets of 25 gm of Food I and Food II purchased. Let Z be the price paid. Obviously, price has to be minimized.

Take a mass balance on the nutrients from Food I and II,

Calcium 10x + 4y ≥ 20

5x + 2y ≥ 10 (i)

Protein 5x + 5 y ≥ 20

x + y ≥ 4 (ii)

Calories 2x + 6y > 13 (iii)

These become the constraints for the cost function, Z to be minimized i.e., 0.6x + y = 7, given cost of Food I is Rs 0.6/ - and Rs 1/ - per lb

From (i), (ii) & (iii) we get points on the X & Y - axis as [0, 5] & [2, 0] ; [0, 4] & [ 4, 0] ; [0, 13/6] & [6.5, 0] Plotting these

The smallest value of Z is 2.9 at the point (2.75, 1.25). We cannot say that the minimum value of z is 2.9 as the feasible region is unbounded.

Therefore, we have to draw the graph of the inequality 0.6x + y <2.9

Plotting this to see if the resulting line (in green) has any point common with the feasible region. Since there are no common points this is the minimum value of the function Z and the mix is

Food I = 2.75 lb; Food II = 1.25 lb; Price = Rs 2.9

When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities., this optimal value must occur at a corner point (vertex) of the feasible region.

Here the feasible region is the unbounded region A - B - C - D

Computing the value of 7 at the corner points of the feasible region ABHG

Let required production of product A and B be x and y respectively

Given, profits on one unit of product A and B are Rs 2 and Rs 3 respectively, so profits on x units of product A and y units of product B are given by 2x and 3y respectively. Let Z be total profit, so

Z = 2x + 3y

Given, production of 1 unit of product A and B require 1 hour and 2 hours of grinding respectively, so, production of x units of product A and y units of product B require x hours and 2y hours of grinding respectively but the maximum time available for grinding is 3 hours, so

x + 2y ≤ 30 (First constraint)

Given, production of 1 unit of product A and B require 3 hours and 1 hours of turning respectively, so x units of product A and y units of product B require 3x hours and y hours of turning respectively but total time available for turning is 60 hours, so

3x + y ≤ 60 (Second constraint)

Given, production of 1 unit of product A and B require 6 hour and 3 hours of assembling respectively, so production of x units of product A and y units of product B require 6x hours and 3y hours of assembling respectively but total time available for assembling is 200 hours, so

6x + 3y ≤ 200 (Third constraint)

Given, production of 1 unit of product A and B require 5 hours and 4 hours of testing respectively, so production of x units of product A and y units of product B require 5x hours and 4y hours of testing respectively but total time available for testing is 200 hours, so

5x + 4y ≤ 200 (Fourth constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

maximize Z = 2x + 3y

Subject to constraints,

x + 2y ≤ 30

3x + y ≤ 60

6x + 3y ≤ 200

5x + 4y ≤ 200

and, x, y ≥ 0 [Since production cannot be negative]

Given information can be tabulated as below:

Let required quantity of food F₁ be x units and quantity of food F₂ be y units.

Given, costs of one unit of food F₁ and F₂ are Rs 5 and Rs 2.5 respectively, so costs of x units of food F₁ and y units of food F₂ are Rs 5x and Rs 2.5y respectively.

Let Z be the total cost, so

Z = 5x + 2.5y

Given, one unit of food F₁ and food F₂ contain 2 and 4 units of vitamin A respectively, so x unit of Food F₁ and y units of food F₂ contain 2x and 4y units of vitamin A respectively, but minimum requirement of vitamin A is 40 unit, so

2x + 4y ≥ 40 (First constraint)

Given, one unit of food F₁ and food F₂ contain 3 and 2 units of vitamin B respectively, so x unit of Food F₁ and y units of food F₂ contain 3x and 2y units of vitamin B respectively, but minimum daily requirement of vitamin B is 40 unit, so

3x + 2y ≥ 50 (Second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

Minimize Z = 5x + 2.5y

Subject to constraint,

2x + 4y ≥ 40

3x + 2y ≥ 50

x, y ≥ 0 [Since requirement of food F₁ and F₂ cannot be less zero.]

Let number of automobiles produces be x and let the number of trucks

Produced be y.

Let Z be the profit function to be maximized.

Z = 2000x + 30000y

The constraints are on the man hours worked

Shop A 2x + 5y ≤ 180 (i) assembly

Shop B 3x + 3y ≤ 135 (ii) finishing

x, y ≥ 0

Corner points can ve obtained from

2x = 3y + 5y = 180 ⇒ x = 0; y = 36 and x = 90; y = 0

3x + 3y ≤ 135 ⇒ x = 0; y = 45 and x = 45; y = 0

Solving (i) and (ii) gives x = 15 and y = 30

Thus 0 automobiles and 36 trucks give max. profit of Rs 10, 80, 000/ -

The data can be represented in a table below

To minimize labour cost means to assume to minimize the earnings i.e,

Min Z = 150x + 200y

With constraints

x, y ≥ 0 at least 1 shirt and 1 pant is required

6x + 10y ≥ 60 require at least 60 shirts

4x + 4y ≥ 32 require at least 32 pants

On solving the above inequalities as equations, we get,

x = 5 and y = 3

other corner points obtained are [0, 6], [10, 0], [0, 8] and [8, 0]

The feasible region is the upper unbounded region A - E - D

Point E(5, 3) may not be minimal value. So, plot 150x + 200y < 1350 to see

If there is a common region with A - E - D.

The green line has no common point, therefore

Thus, stitching 5 shirts and 3 pants minimizes labour cost to Rs 1350/ -

The above LPP can be presented in a table above.

The flight cost is to be minimized i.e; Min Z = x + 1.5y

the constraints

x ≥ 2 at least 2 planes of model 314 must be used

y ≥ 0 at least 1 plane of model .53.5 must be used

20x + 20y ≥ 160 require at least 160 F class seats

30x + 60y ≥ 300 require at least 300 T class seats

Solving the above inequalities as equations we get,

When x = 0, y = 8 and when y = 0, x = 8

When x = 0, y = 5 and when y = 0, x = 10

We get an unbounded region 8 - E - 10 as a feasible solution. Plotting the corner points and evaluating we have,

Since we obtained an unbounded region as the feasible solution a plot of Z (x + 1.5y< 9) is plotted.

Since there are no common points point E is the point that gives a minimum value.

Using 6 planes of model 314 & 2 of model 535 gives minimum cost of 9 lakh rupees.

Let he should solve x, y, z questions from set I, II and III respectively.

Given, each question from set I, II, III earn 5, 4, 6 points respectively, so x questions of set I, y questions of set II and z questions of set III earn 5x, 4y and 6z points, let total point credit be U

So, U = 5x + 4y + 6z

Given, each question of set I, II and III require 3, 2 and 4 minutes respectively, so, x questions of set I, y questions of set II and z questions of set III require 3x, 2y and 4z minutes respectively but given that total time to devote in all three sets is

hours = 210 minutes and first two sets is hours = 150 minutes

So,

3x + 2y + 4z ≤ 210 (First constraint)

3x + 2y ≤ 150 (Second constraint)

Given, total number of questions cannot exceed 100

So, x + y + z ≤ 100 (Third constraint)

Hence, mathematical formulation of LPP is

Find x and y which

maximize U = 5x + 4y + 6z

Subject to constraint,

3x + 2y + 4z ≤ 210

3x + 2y ≤ 150

x + y + z ≤ 100

x, y, z ≤ 0 [Since number of questions to solve from each set

cannot be less than zero.]

Given information can be tabulated below

Let required quantity of field for tomatoes, lettuce and radishes be x, y and z

acre respectively.

Given, costs of cultivation and harvesting of tomatoes, lettuce and radishes are 5 x20 = Rs 100, 6 x20 = Rs 120, 5 x20 = Rs 100 respectively per acre. Cost of fertilizers for tomatoes, lettuce and radishes 100 x 0.50 = Rs 50, 100 x 0.50 = Rs 50 and 50 x0.50 = Rs 25 respectively per acre.

So, total costs of production of tomatoes, lettuce and radishes are (Rs 100 + 50)x = Rs 150x, (Rs 120 + 50)y = Rs 170y and radishes are (Rs 100 + 25)z = Rs 125z respectively total selling price of tomatoes, lettuce and radishes, according to yield are (Rs2000x1)x = Rs 2000x, (Rs3000 x0.75)y = Rs 2250y and (Rs1000x2)z = Rs 2000z respectively.

Let U be the total profit,

So,

U = (2000x - 150x) + (2250y - 170y) + (2000z - 125z)

U = 1850x + 2080y + 1875z

Given, farmer has 100 - acre farm

So, x + y + z ≤ 100 (First constraint)

Number of cultivation and harvesting days are 400 So, 5x + 6y + 5z ≤ 400

Hence, mathematical formulation of LPP is,

Find x, y, z which

maximize U = 1850x + 2080y + 1875z

Subject to constraint,

x + y + z ≤ 100

5x + 6y + 5z ≤ 400

x, y, z ≥ 0 [Since farm used for cultivation cannot be less than zero.]

Given information can be tabulated as below:

Let the required product of product A and B be x and y units respectively.

Given, labour cost and raw material cost of one unit of product A is Rs 16 and Rs 4, so total cost of product A is Rs 16 + Rs 4 = Rs 20

And given selling price of 1 unit of product A is Rs 25,

So, profit on one unit of product

A = 25 - 20 = Rs 5

Again, given labour cost and raw material cost of one unit of product B is Rs 20 and Rs 4 So, that cost of product B is Rs 20 + Rs 4 = Rs 24

And given selling price of 1 unit of product B is Rs 30

So, profit on one unit of product B = 30 - 24 = Rs 6

Hence, profits on x unit of product A and y units of product B are Rs 5x and Rs 6y respectively.

Let Z be the total profit, so,

Z = 5x + 6y

Given, production of one unit of product A and B need to process for 3 and 4 hours respectively in department 1, so production of x units of product A and y units of product B need to process for 3x and 4y hours respectively in Department 1. But total capacity of Department 1 is 130 hours,

So, 3x + 2y ≤ 130 (First constraint)

Given, production of one unit of product. A and B need to process for 4 and 6 hours respectively in department 2, so production of x units of product A and y units of product B need to process for 4x and 6y hours respectively in Department 2 but total capacity of Department 2 is 260 hours

So, 4x + 6y ≤ 260 (Second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

Maximize Z = 5x + 6y

Subject to constraint,

3x + 2y ≤ 130,

4x + 6y ≤ 260

x, y ≥ 0 [Since production cannot be less than zero]

We have to maximize the profit by calculating the number of units needed to produce each product.

For Product A,

Cost Price per unit = Labour Cost + Raw material cost per unit

Cost Price per unit = 16 + 4 = Rs. 20

Selling Price per unit = Rs. 25

Profit per unit = S.P – C.P = 25 – 20 = Rs. 5

For Product B,

Cost Price per unit = Labour Cost + Raw material cost per unit

Cost Price per unit = 20 + 4 = Rs. 24

Selling Price per unit = Rs. 30

Profit per unit = S.P – C.P = 30 – 24 = Rs. 6

Let number of units produced of Product A be x and number of units produced of Product B be y.

Hence, Total Profit = 5 x + 6 y

To Maximize : z = 5 x + 6 y

For Department 1,

3 x + 2 y ≤ 130

For Department 2,

4 x + 6 y ≤ 260

Hence,

Z = 5 x + 6 y

3 x + 2 y ≤ 130

4 x + 6 y ≤ 260

x, y ≥ 0 [Since production cannot be less than zero]

Given,

Objective function is: Z = 5x + 3y

Constraints are:

3x + 5y ≤ 15

5x + 2y ≤ 10

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

3x + 5y≤ 15 → 3x + 5y = 15 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = 5 (5,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin the just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

5x + 2y≤ 10 → 5x + 2y = 10 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 5 (0,5) - - - - first coordinate.

Put, y = 0 ⇒ x = 2 (2,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations.

But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + 5y = 15 and 5x + 2y = 10 gives

Similarly solving 3x + 5y = 15 and x = 0 gives (0, 3)

And solving 5x + 2y = 10 and y = 0 gives (2,0)

And solving x = 0 and y = 0 gives (0,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 5x + 3y

∴ Z at

Z at (0,3) = 5× 0 + 3× 3 = 9

Z at (2,0) = 5× 2 + 3× 0 = 10

Z at (0,0) = 0

We can see that Z is maximum at and max. value is

∴ x = ; y = and maximum the value of Z is

Given,

Objective function is: Z = 9x + 3y

Constraints are:

2x + 3y ≤ 13

3x + y ≤ 5

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + 3y ≤ 13 → 2x + 3y = 13 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,13/3) - - - - first coordinate.

Put, y = 0 ⇒ x = 13/2 (13/2,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin the just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + y≤ 5 → 3x + y = 5 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 5 (0,5) - - - - first coordinate.

Put, y = 0 ⇒ x = 5/3 (5/3,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x – axis

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 2x + 3y = 13 and 3x + y = 5 gives

Similarly solving 2x + 3y = 13 and x = 0 gives (0, 13/3)

And solving 3x + y = 5 and y = 0 gives (5/3,0)

And solving x = 0 and y = 0 gives (0,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 9x + 3y

∴ Z at

Z at (0,13/3) = 9× 0 + 3× 13/3 = 13

Z at (5/3,0) = 9× (5/3) + 3× 0 = 15

Z at (0,0) = 0

We can see that Z is maximum at And max. value is 15

Z is also maximum at (5/3,0) with value = 15

This illustrates that 9x + 3y overlaps with 3x + y = 5 .

∴ Z is maximum at all the points on 3x + y = 5, and max value is 15.

Given,

Objective function is: Z = 18x + 10y

Constraints are:

4x + y ≥ 20

2x + 3y ≥ 30

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

4x + y ≥ 20 → 4x + y = 20 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 20 (0,20) - - - - first coordinate.

Put, y = 0 ⇒ x = 5 (5,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin the just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

2x + 3y ≥ 30 → 2x + 3y = 30 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 10 (0,10) - - - - first coordinate.

Put, y = 0 ⇒ x = 15 (15,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x – axis

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 2x + 3y = 30 and 4x + y = 20 gives (3, 8)

Similarly solving 4x + y = 20 and x = 0 gives (0, 20)

And solving 2x + 3y = 30 and y = 0 gives (15,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 18x + 10y

∴ Z at (3, 8) = 18 × 3 + 8 × 10 = 134

Z at (0,20) = 18 × 0 + 10 × 20 = 200

Z at (15,0) = 18 × (15) + 10 × 0 = 270

Note: It is unbounded so we can’t say directly by seeing that z = 134 is minimum because there might be possibility that some other points from feasible region may result a smaller number.

In such a case minima will not be possible.

So we check this by setting inequation corresponding to the minimum value of Z.

∴ inequation is 18x + 10y < 134

As (0,0) satisfies the inequation, so it will bound its region towards origin and hence will not overlap with the feasible region.

∴ We can say that minima are possible.

We can see that Z is minimum at (3, 8) and min. value is 134

∴ Z is minimum at x = 3 and y = 8 ; and min value is 134.

Given,

Objective function is: Z = 50x + 30y

Constraints are:

2x + y ≤ 18

3x + 2y ≤ 34

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + y ≤ 18 → 2x + y = 18 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 18 (0,18) - - - - first coordinate.

Put, y = 0 ⇒ x = 9 (9,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin the just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + 2y ≤ 34 → 3x + 2y = 34 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 17 (0, 17) - - - - first coordinate.

Put, y = 0 ⇒ x = 34/3 (34/3, 0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x – axis

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + 2y = 34 and 2x + y = 18 gives (2,14)

Similarly solving 3x + 2y = 34 and x = 0 gives (0, 17)

And solving 2x + y = 18 and y = 0 gives (9,0)

And solving x = 0 and y = 0 gives (0,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 50x + 30y

∴ Z at (2,14) = 50 × 2 + 30 × 14 = 520

Z at (0,17) = 50 × 0 + 30 × 17 = 510

Z at (9,0) = 50 × (9) + 30 × 0 = 450

Z at (0,0) = 0

We can see that Z is maximum at (2, 14) and max. value is 520

∴ Z is maximum at x = 2 and y = 14 ; and max value is 520.

Given,

Objective function is: Z = 4x + 3y

Constraints are:

8x + 6y ≤ 48

3x + 4y ≤ 24

x ≤ 5

y ≤ 6

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

8x + 6y ≤ 48 → 8x + 6y = 48 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 8 (0,8) - - - - first coordinate.

Put, y = 0 ⇒ x = 6 (6,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + 4y ≤ 24 → 3x + 4y = 24 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 6 (0,6) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

x = 5 and y = 6 are lines parallel to y - axis and x - axis respectively.

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + 4y = 24 and 8x + 6y = 48 gives

Similarly, solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(0,0) (5,0) (0,6), ,

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 4x + 3y

∴ Z at

Z at (0,6) = 4 × 0 + 3 × 6 = 18

Z at (5,0) = 4 × (5) + 3 × 0 = 20

Z at (0,0) = 0

Z at

We can see that Z is maximum at and max. value is 24

∴ Z is maximum at x = 24/7 and y = 24/7 ; and max value is 24

Also it has a maximum at x = 5 and y = 4/3 with max value = 24

Given,

Objective function is: Z = 15x + 10y

Constraints are:

2x + 3y ≤ 70

3x + 2y ≤ 80

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + 3y ≤ 70 → 2x + 3y = 70 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 70/3 (0,70/3) - - - - first coordinate.

Put, y = 0 ⇒ x = 35 (35,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + 2y ≤ 80 → 3x + 2y = 80 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 40 (0,40) - - - - first coordinate.

Put, y = 0 ⇒ x = 80/3 (80/3,0) - - - - second coordinate

x = 0 is the y-axis and y = 0 is the x-axis

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + 2y = 80 and 2x + 3y = 70 gives (20, 10)

Similarly solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(0,0) (80/3 ,0) (0,70/3),(20,10)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 15x + 10y

∴ Z at (20, 10) = 15 × (20) + 10 × (10) = 400

Z at (0,70/3) = 15 × (70/3) + 10 × 0 = 350

Z at (80/3,0) = 15 × (80/3) + 10 × 0 = 400

Z at (0,0) = 0

Clearly, we can see that Z is maximum at and max. value is 400

∴ Z is maximum at x = 20 and y = 10 ; and max value is 400

Also it has a maximum at x = 80/3 and y = 0 with max value = 400

Given,

Objective function is: Z = 10x + 6y

Constraints are:

2x + 5y ≤ 34

3x + y ≤ 12

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + 5y ≤ 34 → 2x + 5y = 34 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 34/5 (0,34/5) - - - - first coordinate.

Put, y = 0 ⇒ x = 17 (17,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + y ≤ 12 → 3x + y = 12 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 12 (0,12) - - - - first coordinate.

Put, y = 0 ⇒ x = 12/3 = 4 (4,0) - - - - second coordinate

x = 0 is the y-axis and y = 0 is the x-axis

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + y = 12 and 2x + 5y = 34 gives (2, 6)

Similarly solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(0, 0), (4, 0), (0, 34/5), (2, 6)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 10x + 6y

∴ Z at (2, 6) = 10 × (2) + 6 × (6) = 56

Z at (0,34/5) = 10 × (0) + 6 × (34/5) = 204/5

Z at (4,0) = 10 × (4) + 10 × 0 = 40

Z at (0,0) = 0

We can see that Z is maximum at (2,6) and max. value is 56

∴ Z is maximum at x = 2 and y = 16 ; and max value is 56

Given,

Objective function is: Z = 3x + 4y

Constraints are:

2x + 2y ≤ 80

2x + 4y ≤ 120

First convert the given inequations into corresponding equations and plot them:

2x + 2y ≤ 80 → 2x + 2y = 80 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 40 (0,40) - - - - first coordinate.

Put, y = 0 ⇒ x = 40 (40,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in a plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

2x + 4y ≤ 120 → 2x + 4y = 120 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 30 (0,30) - - - - first coordinate.

Put, y = 0 ⇒ x = 60 (60,0) - - - - second coordinate

x = 0 is the y-axis and y = 0 is the x-axis

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 2x + 4y = 120 and 2x + 2y = 80 gives (20, 20)

There are no other corners in the region obtained.

So maxima will occur at (20,20)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 3x + 4y

∴ Z at (20,20) = 3×(20) + 4×(20) = 140

Note: As the region is unbounded, so we need to check whether maxima occurs or not.

For this we define inequation using the optimal function if the solution region of the inequation does not coincide with the feasible region, it means it has a maximum.

∴ inequation is : 3x + 4y > 140

Clearly, from the graph we observe that 3x + 4y>140 does not overlap with the feasible region

∴ Z is maximum at (20,20) and max. value is 140

∴ Z is maximum at x = 20 and y = 20 ; and max value is 140

Given,

Objective function is: Z = 7x + 10y

Constraints are:

x + y ≤ 30000

y ≤ 12000

x ≥ 6000

x, y ≥ 0

x ≥ y

First convert the given inequations into corresponding equations and plot them:

x + y ≤ 30000 → x + y = 30000 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 30000 (0,30000) - - - - first coordinate.

Put, y = 0 ⇒ x = 30000 (30000,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

For a line passing through origin, we put any other coordinate to check the region.

x ≥ y → x - y = 0 (corresponding equation)

Line passing through origin (0,0) and (1,1)

x = 0 is the y - axis and y = 0 is the x - axis

y = 12000 (line parallel to x - axis passing through (0,12000))

x = 6000 (line parallel to x – axis passing through (6000,0))

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 30000 and y = 12000 gives (18000, 12000)

Similarly, solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(18000,12000) , (12000,12000) , (6000,0) , (6000,6000) and (30000,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 7x + 10y

∴ Z at

Z at (12000,12000) = 7 × (12000) + 10 × (12000) = 204000

Z at (6000,0) = 7 × (6000) + 10 × 0 = 42000

Z at (6000,6000) = 7×6000 + 10×6000 = 67000

Z at (30000,0) = 7×30000 + 10×0 = 210000

Z at (0,0) = 0

We can see that Z is maximum at (18000, 20000) and max. value is 246000

∴ Z is maximum at x = 18000 and y = 20000 ; and max value is 246000

Given,

Objective function is: Z = 2x + 4y

Constraints are:

x + y ≥ 8

x + 4y ≥ 12

x ≥ 3

y ≥ 2

First convert the given inequations into corresponding equations and plot them:

x + y ≥ 8 → x + y = 8 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 8 (0,8) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

As, x + 4y ≥ 12 → x + 4y = 12

Put x = 0 ⇒ y = 3 coordinate - - - - - (0,3)

Put y = 0 ⇒ x = 12 coordinate - - - - - - (12,0)

y = 2 (line parallel to x - axis passing through (0,2))

x = 3 (line parallel to x – axis passing through (3,0))

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 8 and x = 3 gives (3, 5)

Similarly, solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(3,5) and (6,2)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 2x + 4y

∴ Z at (3,5) = 2×(3) + 4×(5) = 26

Z at (6,2) = 2 × (6) + 4 × (8) = 20

Note: As the region is unbounded as we can’t say blindly that Z = 24 is maximum because there might be other points in the feasible region that may Make Z even greater.

So we need to check whether Z is maximum or not or Z greater than 24 or not.

For this we define inequation using the optimal function if the solution region of the inequation does not coincide with the feasible region, it means it has a maxima

Inequation : 2x + 4y > 24 or x + 2y > 12

Now we again plot the graph with the constraints and the above inequation

Clearly, x + 2y > 24 has solutions in feasible region.

This proves that the values of Z greater than 24 are possible.

∴ Optimal value of Z is not possible.

Given,

Objective function is: Z = 5x + 3y

Constraints are:

2x + y ≥ 10

x + 3y ≥ 15

x ≤ 10

y ≤ 8

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + y ≥ 10 → 2x + y = 10 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 10 (0,10) - - - - first coordinate.

Put, y = 0 ⇒ x = 5 (5,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

For a line passing through origin, we put any other coordinate to check the region.

x + 3y ≥ 15 → x + 3y = 15 (corresponding equation)

Put x = 0 ⇒ y = 5 coordinate - - - - (0,5)

Put y = 0 ⇒ x = 15 coordinate - - - - (15,0)

x = 0 is the y - axis and y = 0 is the x - axis

y = 8 (line parallel to x - axis passing through (0,8))

x = 10 (line parallel to x – axis passing through (10,0))

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + 3y = 15 and 2x + y = 10 gives (3,4)

Similarly solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(3,4) , (10,8), (1,8) and (10, 5/3)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 5x + 3y

∴ Z at (3,4) = 5×(3) + 3×(4) = 27

Z at (10,8) = 5 × (10) + 3 × (8) = 74

Z at (1,8) = 5 × (1) + 3 × 8 = 29

Z at (10,5/3) = 5 × 10 + 3 × 5/3 = 55

We can see that Z is minimum at (3,4) and min. value is 27

∴ Z is minimum at x = 3 and y = 4 ; and min value is 27

Given,

Objective function is: Z = 30x + 20y

Constraints are:

x + y ≤ 8

x + 4y ≥ 12

5x + 8y = 20

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x + y ≤ 8 → x + y = 8 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 8 (0,8) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

For a line passing through origin, we put any other coordinate to check the region.

x + 4y ≥ 12 → x + 4y = 12 (corresponding equation)

Put x = 0 ⇒ y = 3 coordinate - - - - (0,3)

Put y = 0 ⇒ x = 12 coordinate - - - - (12,0)

x = 0 is the y - axis and y = 0 is the x - axis

5x + 8y = 20

On putting x = 0 , y = 20/8 (0,2.5)

Putting y = 0 , x = 4 (4,0)

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 8 and x + 4y = 12 gives (20/3,4/3)

Similarly, solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(0,3) , (0,8) and (20/3 , 4/3)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 30x + 20y

∴ Z at (0,8) = 30×(0) + 20×(8) = 160

Z at (0,3) = 30 × (0) + 20 × (3) = 60

Z at (20/3,4/3) = 30 × (20/3) + 20 × 4/3 = 226.666We can see that Z is minimum at (0,3) and min. value is 60

∴ Z is minimum at x = 0 and y = 3 ; and min value is 60

Maximum value is 226.66 and it occurs at x = 20/3 and y = 4/3

Ideas required to solve the problem:

• Fundamentals of plotting a linear equation. 2 coordinates are sufficient to plot a straight line.

• A linear inequation represents a region of XY plane when plotted.

• For linear programming we define various linear constraints and combining them we get a region in the XY plane which represents a region of feasible operation subject to various constraints.

• But our objective in the linear programming problem is to optimize (maximize or minimize) our objective function and it will be optimal only at one of the corner points of the feasible region. So we check the value of the objective function at every corner points and hence find maximum or minimum

Given,

Objective function is: Z = 4x + 3y

Constraints are:

8x + 6y ≤ 48

3x + 4y ≤ 24

x ≤ 5

y ≤ 6

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

8x + 6y ≤ 48 → 8x + 6y = 48 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 8 (0,8) - - - - first coordinate.

Put, y = 0 ⇒ x = 6 (6,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + 4y ≤ 24 → 3x + 4y = 24 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 6 (0,6) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

x = 5 and y = 6 are lines parallel to y - axis and x - axis respectively.

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 3x + 4y = 24 and 8x + 6y = 48 gives

Similarly, solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(0,0) (5,0) (0,6), ,

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 4x + 3y

∴ Z at

Z at (0,6) = 4 × 0 + 3 × 6 = 18

Z at (5,0) = 4 × (5) + 3 × 0 = 20

Z at (0,0) = 0

Z at

We can see that Z is maximum at and max. value is 24

∴ Z is maximum at x = 24/7 and y = 24/7 ; and max value is 24

Also it has a maximum at x = 5 and y = 4/3 with max value = 24

Given,

Objective function is: Z = x - 5y + 20

Constraints are:

x - y ≥ 0

- x + 2y ≥ 2

x ≥ 3

y ≤ 4

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x - y ≥ 0 → x = y (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 0 (0,0) - - - - first coordinate.

Put, y = 1 ⇒ x = 1 (1,1) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

- x + 2y ≥ 2 → - x + 2y = 2 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 1 (0,1) - - - - first coordinate.

Put, y = 0 ⇒ x = - 2 ( - 2,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

x = 3 and y = 4 are lines parallel to y - axis and x - axis respectively.

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x - y = 0 and x = 3 gives (3,3)

Similarly, solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(3,3) (4,4) (6,4),

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = x - 5y + 20

∴ Z at

Z at (3,3) = 3 – 5×3 + 20 = 8

Z at (4,4) = 4 - 5×4 + 20 = 4

Z at (6,4) = 6 – 5×4 + 20 = 6

We can see that Z is minimum at (4,4) and min. value is 4

∴ Z is minimum at x = 4 and y = 4 ; and min value is 4

Given,

Objective function is: Z = 3x + 5y

Constraints are:

x + 2y ≤ 20

x + y ≤ 15

y ≤ 5

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x + 2y ≤ 20 → x + 2y = 20 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 10 (0,10) - - - - first coordinate.

Put, y = 0 ⇒ x = 20 (20,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + y ≤ 15 → x + y = 15 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 15 (0,15) - - - - first coordinate.

Put, y = 0 ⇒ x = 15 (15,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

y = 5 is line parallel to x - axis passing through (0,5).

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 15 and x + 2y = 20 gives (10,5)

Similarly solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(0,0) (10,5) (15,0), (0,5)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 3x + 5y

∴ Z at (0,0) = 0

Z at (10,5) = 3× 10 + 5×5 = 55

Z at (15,0) = 3×15 + 5× 0 = 45

Z at (0,5) = 3×0 + 5×5 = 25

We can see that Z is maximum at (10,5) and max. value is 55

∴ Z is maximum at x = 10 and y = 5 ; and max value is 55

Given,

Objective function is: Z = 3x₁ + 5x₂

Constraints are:

x₁ + 3x₂ ≥ 3

x₁ + x₂ ≥ 2

x₁, x₂ ≥ 0

First convert the given inequations into corresponding equations and plot them:

x₁ + 3x₂ ≥ 3 → x₁ + 3x₂ = 3 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 1 (0,1) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = 3 (3,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x₁ + x₂ ≥ 2 → x₁ + x₂ = 2 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 2 (0,2) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = 2 (2,0) - - - - second coordinate

x₁ = 0 is the y-axis and x₂ = 0 is the x-axis

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x₁ + x₂ = 2 and x₁ + 3x₂ = 3 gives (3/2,1/2)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(3/2,1/2) ,(3,0) and (0,2)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 3x₁ + 5x₂

∴ Z at (3,0) = 3× 3 + 5× 0 = 9

Z at (0,2) = 3× 0 + 5×2 = 10

Z at (3/2,1/2) = 3×(3/2) + 5×(1/2) = 7

Note: As the region is unbounded, so we need to check whether any value less than 7 is possible for Z or not. If it is unique, we will say that under given constraints we found the minimum Z

For this we define inequation as: 3x + 2y < 7

As (0,0) satisfies the inequation, so this is the region specified by above inequation. As our feasible region does not coincide with the region specified by 3x + 2y < 7.

∴ Z can be minimized

We can see that Z is minimum at (3/2,1/2) and min. value is 7

∴ Z is minimum at x = 3/2 and y = 1/2 ; and min value is 7

Given,

Objective function is: Z = 2x + 3y

Constraints are:

x + y ≥ 1

10x + y ≥ 5

x + 10y ≥ 1

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x + y ≥ 1 → x + y = 1 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 1 (0,1) - - - - first coordinate.

Put, y = 0 ⇒ x = 1 (1,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

10x + y ≥ 5 → 10x + y = 5 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 5 (0,5) - - - - first coordinate.

Put, y = 0 ⇒ x = 1/2 (1/2,0) - - - - second coordinate

x + 10y ≥ 1 → x + 10y = 1 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 1/10 (0,0.1) - - - - first coordinate.

Put, y = 0 ⇒ x = 1 (1,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 1 and 10x + y = 5 gives (4/9,5/9)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(4/9,5/9) (1,0) and (0,5)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 2x + 3y

∴ Z at (1,0) = 2

Z at (0,5) = 2× 0 + 3×5 = 15

Z at (4/9,5/9) = 2×(4/9) + 3× (5/9) = 23/9

Note: As the region is unbounded as we can’t say blindly that Z = 15 is maximum because there might be other points in the feasible region that may Make Z even greater.

So we need to check whether Z is maximum or not or Z greater than 15 or not.

For this we define inequation using the optimal function if the solution region of the inequation does not coincide with the feasible region, it means it has a maxima

Inequation : 2x + 3y > 15

We can see that the inequation 2x + 3y > 15 overlaps with a feasible region. So there are no maxima possible

If we want to check regarding a minimum of Z, from the corner we have a min of Z = 2

∴ required inequation is : 2x + 3y<2

Clearly no overlap with the feasible region is there for 2x + 3y < 2

∴ Z has minimum at x = 1 and y = 0 and min value of Z is 2

Given,

Objective function is: Z = - x₁ + 2x₂

Constraints are:

- x₁ + 3x₂ ≤ 10

x₁ + x₂ ≤ 6

x₁ - x₂ ≤ 2

x₁, x₂ ≥ 0

First convert the given inequations into corresponding equations and plot them:

- x₁ + 3x₂ ≤ 10 → - x₁ + 3x₂ = 10 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 10/3 (0,10/3) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = - 10 ( - 10,0) - - - - second coordinate

x₁ + x₂ ≤ 6 → x₁ + x₂ = 6 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 6 (0,6) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = 6 (6,0) - - - - second coordinate

x₁ - x₂ ≤ 2 → x₁ - x₂ = 2 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = - 2 (0, - 2) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = 2 (2,0) - - - - second coordinate

x₁ ≥ 0 and x₂ ≥ 0

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

Hence, we obtain the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x₁ - x₂ = 2 and x₁ + x₂ = 6 gives (6,6)

Similarly solve other combinations by observing graph to get other coordinates.

From figure we have obtained coordinates of corners as:

(0,0),(2,0),(4,2),(2,4) and (0,10/3)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = - x₁ + 2x₂

∴ Z at (0,0) = 0

Z at (2,0) = - 2 + 2×0 = - 2

Z at (4,2) = - 4 + 2× 2 = 0

Z at (2,4) = - 2 + 2×4 = 6

Z at (0,10/3) = - 0 + 2× (10/3) = 20/3

We can see that Z is maximum at (0,10/3) and max. value is 20/3

∴ Z is maximum at x = 0 and y = 10/3 ; and max value is 20/3

Given,

Objective function is: Z = x + y

Constraints are:

- 2x + y ≤ 1

x ≤ 2

x + y ≤ 3

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

- 2x + y ≤ 1 → - 2x + y = 1 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 1 (0,1) - - - - first coordinate.

Put, y = 0 ⇒ x = - 1/2 ( - 1/2,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + y ≤ 3 → x + y = 3 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = 3 (3,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis

x = 2 is line parallel to y - axis passing through (2,0).

Hence we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving - 2x + y = 1 and x + y = 3 gives (2/3,7/3)

Similarly, solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(2/3,7/3), (0,1), (0,0), (2,0) and (2,1)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = x + y

∴ Z at (0,0) = 0

Z at (2,0) = 2 + 0 = 2

Z at (2,1) = 2 + 1 = 3

Z at (0,1) = 0 + 1 = 1

Z at (2/3,7/3) = 2/3 + 7/3 = 3

Clearly, we can see that Z is maximum at (2,1) and max. value is 3

∴ Z is maximum at x = 2 and y = 1 also at x = 2/3 and y = 7/3; and max value is 3

Given,

Objective function is: Z = 3x₁ + 4x₂

Constraints are:

x₁ - x₂ ≤ - 1

- x₁ + x₂ ≤ 0

x₁ , x₂ ≥ 0

First convert the given inequations into corresponding equations and plot them:

x₁ - x₂ ≤ - 1 → x₁ - x₂ = - 1 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 1 (0,1) - - - - first coordinate.

Put, x₂ = 0 ⇒ x₁ = - 1 ( - 1,0) - - - - second coordinate

- x₁ + x₂ ≤ 0 → - x₁ + x₂ = 0 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x₁ = 0 ⇒ x₂ = 0 (0,0) - - - - first coordinate.

Put, x₂ = 1 ⇒ x₁ = - 1 ( - 1,1) - - - - second coordinate

x₁ ≥ 0 and x₂ ≥ 0

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

Hence, we obtain the following plot:

The given constraints don’t enclose any feasible region. No common shaded region so no maxima exists.

Given,

Objective function is: Z = 3x + 4y

Constraints are:

x – y ≤ 1

x + y ≥ 3

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x – y ≤ 1 → x – y = 1 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = - 1 (0, - 1) - - - - first coordinate.

Put, y = 0 ⇒ x = 1 (1,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + y ≤ 3 → x + y = 3 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = 3 (3,0) - - - - second coordinate

x = 0 is the y-axis and y = 0 is the x-axis

Hence, we obtain a plot as shown in figure:

The shaded region in the above figure represents the region of a feasible solution.

Now to maximize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 3 and x = 0 gives (0,3)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(0,3) and (3,0)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 3x + 4y

∴ Z at (0,3) = 4×3 = 12

Z at (3,0) = 3×3 + 4×0 = 9

As the region is unbounded as we can’t say blindly that Z = 12 is maximum because there might be other points in the feasible region that may Make Z even greater.

So we need to check whether Z is maximum or not or Z greater than 12 or not.

For this we define inequation using the optimal function if the solution region of the inequation does not coincide with the feasible region, it means it has a maxima

Inequation: 3x + 4y > 12

It will be away from the origin side so that it will overlap with the feasible region.

∴ No maxima are possible.

Given,

Z = 3x + 2y

Constraints:

5x + y ≥ 10

x + y ≥ 6

x + 4y ≥ 12

x ≥ 0, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

5x + y ≥ 10 → 5x + y = 10(corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 10 (0,10) - - - - first coordinate.

Put, y = 0 ⇒ x = 2 (2,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + y ≥ 6 → x + y = 6 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 6 (0,6) - - - - first coordinate.

Put, y = 0 ⇒ x = 6 (6,0) - - - - second coordinate

x + 4y ≥ 12 → x + 4y = 12 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = 12 (12,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis.

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + 4y = 12 and x + y = 6 gives (4,2)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(4,2),(12,0),(1,5) and (0,10)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 3x + 2y

∴ Z at (4,2) = 4×3 + 2×2 = 16

Z at (12,0) = 3×12 + 2×0 = 36

Z at (1,5) = 3× 1 + 2× 5 = 13

Z at (0,10) = 3× 0 + 2×10 = 20

As the region is unbounded as we can’t say blindly that Z = 13 is minimum because there might be other points in feasible region that may Make Z even lesser.

So we need to check whether Z is minimum or not or Z lesser than 13 or not.

For this we define inequation using the optimal function if the solution region of the inequation does not coincide with the feasible region, it means it has a maxima

Inequation: 3x + 2y < 13

It will be towards origin side, so it will not overlap with the feasible region.

∴ Minima are possible.

Minimum is possible and minimum occurs at x = 1 and y = 5 and value is Z = 13.

Given,

Z = 2x + y

Constraints:

x + 3y ≥ 6

x – 3y ≤ 3

3x + 4y ≤ 24

- 3x + 2y ≤ 6,

5x + y ≥ 5,

x, y ≥ 0.

First convert the given inequations into corresponding equations and plot them:

x + 3y ≥ 6 → x + 3y = 6 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 2 (0,2) - - - - first coordinate.

Put, y = 0 ⇒ x = 6 (6,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x – 3y ≤ 3 → x – 3y = 3 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = - 1 (0, - 1) - - - - first coordinate.

Put, y = 0 ⇒ x = 3 (3,0) - - - - second coordinate

3x + 4y ≤ 24 → 3x + 4y = 24 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 6 (0,6) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

- 3x + 2y ≤ 6 → - 3x + 2y = 6 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = - 2 ( - 2,0) - - - - second coordinate

5x + y ≥ 5 → 5x + y = 5, (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 5 (0,5) - - - - first coordinate.

Put, y = 0 ⇒ x = 1 (1,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis.

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 5x + y = 5 and - 3x + 2y = 6 gives (4/13,45/13)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(4/13, 45/13) ,(4.5, 0.5), (9/14, 25/14), (4/3, 5), (84/13, 15/13)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 2x + y

∴ Z at = 2×(4/13) + 45/13 = 57/13 = 4.38

Z at = 2×(9/2) + 1/2 = 19/2 = 9.5

Z at (9/14,25/14) = 2× (9/14) + (25/14) = 43/14 = 3.07

Z at = 2×(4/3) + 5 = 23/3 = 7.67

Z at 183/13 = 14.07

Clearly, Z is maximum at x = 84/13 and y = 15/13 and maximum value is 14.07

Z is minimum at x = 9/14 and y = 25/14 and minimum value is 3.07

Given,

Z = 3x + 5y

Constraints:

- 2x + y ≤ 4

x + y ≥ 3

x – 2y ≤ 2

x, y ≤ 0

First convert the given inequations into corresponding equations and plot them:

- 2x + y ≤ 4 → - 2x + y = 4 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 4 (0,4) - - - - first coordinate.

Put, y = 0 ⇒ x = - 2 ( - 2,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + y ≥ 3 → x + y = 3 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 3 (0,3) - - - - first coordinate.

Put, y = 0 ⇒ x = 3 (3,0) - - - - second coordinate

x – 2y ≤ 2 → x – 2y = 2 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = - 1 (0, - 1) - - - - first coordinate.

Put, y = 0 ⇒ x = 2 ( 2,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis.

Hence, we have the following plot:

There is no shaded region in the above figure represents that there is no region of a feasible solution.

x + y ≥ 3 will not be bounded by x, y ≤ 0. Thus, no feasible region is there.

∴ There is no possible minimum value Z.

Given,

Z = 60x + 15y

Constraints:

x + y ≤ 50

3x + y ≤ 90

x, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

x + y ≤ 50 → x + y = 50 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 50 (0,50) - - - - first coordinate.

Put, y = 0 ⇒ x = 50 (50,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + y ≤ 90 → 3x + y = 90 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 90 (0,90) - - - - first coordinate.

Put, y = 0 ⇒ x = 30 (30,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis.

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + y = 50 and 3x + y = 90 gives (20,30)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(0,0),(30,0),(0,50) and (20,30)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 60x + 15y

∴ Z at (0,0) = 0

Z at (30,0) = 60×30 + 15×0 = 1800

Z at (0,50) = 60×0 + 15×50 = 750

Z at (20,30) = 60×20 + 15×30 = 1650

Clearly Z is maximum at x = 30 and y = 0 and maximum value is 1800

Given,

Z = 2x + 5y

Constraints:

2x + 4y ≤ 8

3x + y ≤ 6

x + y ≤ 4

x ≥ 0, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

2x + 4y ≤ 8 → 2x + 4y = 8 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 2 (0,2) - - - - first coordinate.

Put, y = 0 ⇒ x = 4 (4,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + y ≤ 6 → 3x + y = 6 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 6 (0,6) - - - - first coordinate.

Put, y = 0 ⇒ x = 2 (2,0) - - - - second coordinate

x + y ≤ 4 → x + y = 4 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 4 (0,4) - - - - first coordinate.

Put, y = 0 ⇒ x = 4 (4,0) - - - - second coordinate

x = 0 is the y - axis and y = 0 is the x - axis.

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving 2x + 4y = 8 and 3x + y = 6 gives (8/5,6/5)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(0,0),(2,0),(0,2) and (8/5,6/5)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 2x + 5y

∴ Z at (0,0) = 0

Z at (2,0) = 2×2 + 5×0 = 4

Z at (0,2) = 2×0 + 5×2 = 10

Z at (8/5,6/5) = 2×(8/5) + 5×(6/5) = 46/5 = 9.2

Clearly Z is maximum at x = 0 and y = 2 and maximum value is 10

Given,

Z = 20x + 10y

Constraints:

x + 2y ≤ 28

3x + y ≤ 24

x ≥ 2

First convert the given inequations into corresponding equations and plot them:

x + 2y ≤ 28 → x + 2y = 28 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 14 (0,14) - - - - first coordinate.

Put, y = 0 ⇒ x = 28 (28,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

3x + y ≤ 24 → 3x + y = 24 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 24 (0,24) - - - - first coordinate.

Put, y = 0 ⇒ x = 8 (8,0) - - - - second coordinate

x = 2 is the line parallel to y-axis passing through (2,0)

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + 2y = 28 and 3x + y = 24 gives (4,12)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(2,0),(8,0),(2,13) and (4,12)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 20x + 10y

∴ Z at (2,0) = 20× 2 + 10× 0 = 40

Z at (8,0) = 20×8 + 10×0 = 160

Z at (2,13) = 20×2 + 10×13 = 170

Z at (4,12) = 20×(4) + 10×(12) = 200

Clearly Z is maximum at x = 4 and y = 12 and maximum value is 200

Given,

z = 6x + 3y

Constraints:

4x + y ≥ 80

x + 5y ≥ 115

3x + 2y ≤ 150

x ≥ 0, y ≥ 0

First convert the given inequations into corresponding equations and plot them:

4x + y ≥ 80 → 4x + y = 80 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 80 (0,80) - - - - first coordinate.

Put, y = 0 ⇒ x = 20 (20,0) - - - - second coordinate

Join them to get the line.

As we know, Linear inequation will be a region in the plane, and we observe that the equation divides the XY plane into 2 halves only, so we need to check which region represents the given inequation,

If the given line does not pass through origin then just put (0,0) to check whether inequation is satisfied or not. If it satisfies the inequation origin side is the required region else the other side is the solution.

Similarly, we repeat the steps for other inequation also and find the common region.

x + 5y ≥ 115 → x + 5y = 115 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 23 (0,23) - - - - first coordinate.

Put, y = 0 ⇒ x = 115 (115,0) - - - - second coordinate

3x + 2y ≤ 150 3x + 2y = 150 (corresponding equation)

Two coordinates required to plot the equation are obtained as:

Put, x = 0 ⇒ y = 75 (0,75) - - - - first coordinate.

Put, y = 0 ⇒ x = 50 (50,0) - - - - second coordinate

x = 0 is y - axis and y = 0 is the x – axis.

Hence, we have the following plot:

The shaded region in the above figure represents the region of a feasible solution.

Now to minimize our objective function, we need to find the coordinates of the corner points of the shaded region.

We can determine the coordinates graphically our by solving equations. But choose only those equations to solve which gives one of the corner coordinates of the feasible region.

Solving x + 5y = 115 and 3x + 2y = 150 gives (40,15)

Similarly solve other combinations by observing graph to get other coordinates.

From the figure we have obtained coordinates of corners as:

(15,20),(40,15) and (2,72)

Now we have coordinates of the corner points so we will put them one by one to our objective function and will find at which point it is maximum.

∵ Z = 6x + 3y

∴ Z at (15,20) = 6× 15 + 3×20 = 150

Z at (40,15) = 6×40 + 3×15 = 285

Z at (2,72) = 6×2 + 10×72 = 732

Z is minimum at x = 15, and y = 20 and the minimum value is 150

The information above can be expressed in the following table:

Let the amount of food F1 and F2 required to be ‘x’ and ‘y’ units.

Cost of F1 = 0.20x

Cost of F2 = 0.15y

So, Cost of diet = 0.20x + 0.15y

Now,

⟹ 0.25x + 0.10y ≥ 1.00

i.e. the minimum requirement of Thiamine should be 1.00mg, from both the foods F1 and F2, each of which have 0.25mg and 0.10mg of Thiamine respectively. So, this is the first constraint.

⟹ 0.75x +1.50y ≥ 7.50

i.e. the minimum requirement of Phosphorous should be 7.50mg, from both the foods F1 and F2, each of which have 0.75mg and 1.50mg of Phosphorous respectively. This is the second constraint.

⟹ 1.60x + 0.80y ≥ 10.00

i.e. the minimum requirement of Iron should be 10.00mg, from both the foods F1 and F2, each of which have 1.60mg and 0.8mg of Iron respectively.

Hence, mathematical formulation of LPP is as follows:

Find ‘x’ and ‘y’ which

Minimise Z = 0.20x + 0.15y

Subject to the following constraints:

(i) 0.25x + 0.10y ≥ 1.00

(ii) 0.75x +1.50y ≥ 7.50

(iii) 1.60x + 0.80y ≥ 10.00

(iv) x,y ≥0 (∵ quantity cant be negative)

The feasible region is unbounded

The end points of the feasible region are as follows:

So, Z is smallest at B(5,2.5)

Let us consider the inequation 0.20x + 0.15y ≤ 1.375

As this has no intersection with the feasible region, the smallest value is the minimum value.

So, the minimum cost of diet is ₹1.375

The above information can be expressed in the form of the following table:

Let the quantity of the foods be ‘x’ and ‘y’ respectively.

Cost of food A = 4x

Cost of food B = 3y

Total cost of the combination = 4x + 3y

Now,

⟹ 200x + 100y ≥ 4000

i.e. the minimum requirement of vitamins from the two foods should be 4000.

⟹ x + 2y ≥ 50

i.e. the minimum requirement of minerals from the two foods should be 50.

⟹ 40x + 40y ≥ 1400

i.e. the minimum requirement of calories from the two foods should be 1400

Hence, mathematical formulation of LPP is as follows:

Find ‘x’ and ‘y’ which

Minimize Z = 4x + 3y

Subject to the following constraints:

(i) 200x + 100y ≥ 4000

i.e. 2x + y ≥ 40

(ii) x + 2y ≥ 50

(iii) 40x + 40y ≥ 1400

i.e. x + y ≥ 35

(iv) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region is as follows:

Z is smallest at B(5,30)

Let us consider 4x + 3y ≤ 110

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of foods is ₹110

Let the quantity of foods chosen be ‘x’ and ‘y’

Cost of food X = 0.6x

Cost of food Y = y

Cost of diet = 0.6x + y

Now,

⟹ 10x + 4y ≥ 20

i.e. the minimum daily requirement of calcium in the diet is 20 units.

⟹ 5x + 6y ≥ 20

i.e. the minimum daily requirement of protein in the diet is 20 units.

⟹ 2x + 6y ≥ 12

i.e. the minimum daily requirement of calories in the diet is 12 units.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and y’ such that

Minimises Z = 0.6x + y

Subject to the following constraints:

(i) 10x + 4y ≥ 20

(ii) 5x + 6y ≥ 20

(iii) 2x + 6y ≥ 12

(iv) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded

The corner points of the feasible region is as follows:

Z is smallest at

Let us consider 0.6x + y ≤ 2.712.

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum value of Z is ₹2.712

The above information can be expressed using the following table:

Let the quantity of the foods A and B be ‘x’ and ‘y’ respectively.

Cost of food A = 0.10x

Cost of food B = 0.10y

Cost of diet = 0.10x + 0.10y

Now,

⟹ 0.12x + 0.10y ≥ 0.5

i.e. the minimum requirement of thiamine in the foods is 0.5mg

⟹ 100x + 150y ≥ 600

i.e. the minimum requirement of calories in the foods is 600.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that:

Minimises Z = 0.10x + 0.10y

Subject to the following constraints:

(i) 0.12x + 0.10y ≥ 0.5

(ii) 100x + 150y ≥ 600

i.e. 2x + 3y ≥ 12

(iii) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region is as follows:

Z is smallest at A(1.875,2.75)

Let us consider 0.1x+0.1y ≤ 0.4625

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of the foods is ₹0.4625

The above information can be expressed with the help of the following table:

Let the quantity of foods X and Y be ‘x’ and ‘y’.

Cost of food X = 5x

Cost of food Y = 8y

Cost of the meal 5x+8y

Now,

⟹ x + 2y ≥ 6

i.e. the minimum requirement of Vitamin A in the foods X and Y is 6units, each of which has 1unit and 2 unit of Vitamin A.

⟹ x + y ≥ 7

i.e. the minimum requirement of Vitamin B in the two foods is 7units, each of which has 1 unit of Vitamin B.

⟹ x + 3y ≥ 11

i.e. the minimum requirement of vitamin C in the two foods is 11units, each of which has 1 unit and 3 units of vitamin C.

⟹ 2x + y ≥ 9

i.e. the minimum requirement of Vitamin D in the foods is 9units, each of which has 2 units and 1 unit of Vitamin D.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that

Minimises Z = 5x + 8y

Subject to the following constraints:

(i) x + 2y ≥ 6

(ii) x + y ≥ 7

(iii) x + 3y ≥ 11

(iv) 2x + y ≥ 9

(v) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region are as follows:

Z is smallest at C(5,2)

Let us consider 5x + 8y ≤ 41.

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of the diet is ₹41

The above information can be expressed in the following table:

Let the quantity of the foods F1 and F2 be ‘x’ and ‘y’ respectively.

Cost of food F1 = 4x

Cost of food F2 = 6y

Cost of Diet = 4x + 6y

Now,

⟹ 3x + 6y ≥ 80

i.e. the minimum requirement of Vitamins from the two foods is 80units, each of which contains 3units and 6units of Vitamins.

⟹ 4x + 3y ≥ 100

i.e. the minimum requirement of minerals firm the two foods is 100units, each of which contains 4unit and 3 units of vitamins.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that:

Minimise Z = 4x + 6y

Subject to the following constraints:

(i) 3x + 6y ≥ 80

(ii) 4x + 3y ≥ 100

(iii) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region is as follows:

Z is smallest at

Let us consider 4x + 6y ≤ 104

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of diet is ₹104.

The above information can be expressed using the following table:

Let the amount of Bran and Rice required be ‘x’ and ‘y’ kgs respectively.

Cost of Bran = 5x

Cost of Rice = 4y

Cost of the cereal = 5x + 4y

Now,

⟹ 80x + 100y ≥ 88

i.e. the minimum requirement of protein in the cereal, from Bran and Rice combined, is 88g, each of which have 80g and 100g of proteins respectively.

⟹ 40x + 30y ≥ 36

i.e. the minimum requirement of iron in the cereal, from Bran and Rice combined, is 36mg, each of which have 40mg and 30mg of iron.

Hence, mathematical formulation of LPP is as follows:

Find ‘x’ and ‘y’ that:

Minimises Z = 5x + 4y

Subject to the following constraints:

(i) 80x + 100y ≥ 88

(ii) 40x + 30y ≥ 36

(iii) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region is as follows:

Z is smallest at B(0.6,0.4)

Let us consider 5x + 4y ≤ 4.6

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of the cereal is ₹4.6

The above information can be expressed in the form of the following table:

Let the number of bags chosen of A and B be ‘x’ and ‘y’ respectively.

Cost of Bag A = 8x

Cost of Bag B = 12y

Total Cost of Bags = 8x + 12 y

Now,

⟹ 60x + 30y ≥ 240

i.e. the minimum requirement of almonds from both the bags is 240g, each of which contains 60g and 30g of almonds respectively.

⟹ 30x + 60y ≥ 300

i.e. the minimum requirement of Cashew Nuts from both the bags is 300g, each of which contains 30g and 60g of cashew nuts respectively.

⟹ 30x + 180y ≥ 540

i.e. the minimum requirement of Hazel Nuts from both the bags is 540g, each of which contains 30g and 180g of hazelnut respectively.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that

Minimises Z = 8x + 12y

Subject to the following constraints:

(i) 60x + 30y ≥ 240

i.e. 2x + y ≥ 8

(ii) 30x + 60y ≥ 300

i.e. x + 2y ≥ 10

(iii) 30x + 180y ≥ 540

i.e. x + 6y ≥ 18

(iv) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region are as follows:

Z is smallest at C(2,4)

Let us consider 8x + 12y ≤ 64

As this has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of the bags is ₹64

The above information can be expressed in the form of the following table

Let ‘x’ and ‘y’ units of cake 1 and cake 2 be made.

Number of cakes made = x + y

Now,

⟹ 300x + 150y ≤ 7500

i.e. the maximum availability of flour is 7500g for both cakes, each of which requires 300g and 150g of flour respectively

⟹ 15x + 30y ≥ 600

i.e. the maximum availability of fat is 600g for both the cakes, each of which requires 15g and 30g of fat.

Hence , mathematical formulation of the LPP is as follows :

Find ‘x’ and ‘y’ that,

Maximises Z = x + y

Subject to the following constraints:

(i) 300x + 150y ≤ 7500

i.e. 2x + y ≤ 50

(ii) 15x + 30y ≥ 600

i.e. x + 2y ≥ 40

(iii) x,y ≥0 (∵ quantity cant be negative)

The feasible region is bounded (ABO)

The corner points of the feasible region is as follows:

Z is maximised at B(20,10)

The maximum number of cakes that can be made are 20 and 10 of each kind i.e. 30 in total.

The above information can be expressed in the form of the following table:

Let the mixture contain ‘x’ kgs and ‘y’ kgs of food P and Q respectively.

Cost of food P = 60x

Cost of food Q = 80y

Cost of mixture = 60x + 80y

Now,

⟹ 3x + 4y ≥ 8

i.e. the minimum requirement of vitamin A from the mixture of P and Q is 8units, each of which contains 3units and 4units respectively.

⟹ 5x + 2y ≥ 11

i.e. the minimum requirement of vitamin B from the mixture of P and Q is 11 units, each of which contains 5units and 2units respectively.

Hence, mathematical formulation of the LPP is as follows:

The feasible region is Unbounded.

The corner points of the feasible region are as follows:

Z is minimised on the line joining points B(2,0.5) and .

The minimum cost of mixture is ₹160

The above information can be expressed in the form of the following table:

Let the number of Cake 1 and Cake 2 be made be ‘x’ and ‘y’

Number of cakes made = x + y

Now,

⟹ 200x + 100y ≤ 5000

i.e. the maximum flour available for both the cakes combined is 5000g, each of which requires 200g and 100g of flour respectively.

⟹ 25x + 50y ≤ 1000

i.e. the maximum fat available for the two cakes combined is 1000g, each of which requires 25g and 50g of fat respectively.

Hence, the mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that:

Maximises Z = x + y

Subject to the following constraints:

(i) 200x + 100y ≤ 5000

i.e. 2x + y ≤ 50

(ii) 25x + 50y ≤ 1000

i.e. x + 2y ≤ 40

(iii) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is bounded (OBAC)

The corner points of the feasible region is as follows:

Z is maximised at A(20,10)

The maximum number of cakes that can be made are 30.

The above information can be expressed with the help of the following table

Let the number of packets bought, of P and Q, be ‘x’ and ‘y’

Vitamin A from P = 6x

Vitamin A from Q = 3y

Vitamin A in the diet = 6x + 3y

Now,

⟹ 12x + 3y ≥ 240

i.e. the minimum requirement of Calcium in the diet, form both the foods combined, is 240units, each of which has 12units and 3units of calcium respectively.

⟹ 4x + 20y ≥ 460

i.e. the minimum requirement of Iron from P and Q combined is 460 units, each of which has 4units and 20units of iron respectively.

⟹ 6x + 4y ≤ 300

i.e. the maximum requirement of Cholesterol from P and Q combined is 300 units, each of which contains 6units and 4units of cholesterol respectively.

Hence, the mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that:

Minimises Z = 6x + 3y

Subject to the following constraints:

(i) 12x + 3y ≥ 240

i.e. 4x + y ≥ 80

(ii) 4x + 20y ≥ 460

i.e. x + 5y ≥ 115

(iii) 6x + 4y ≤ 300

i.e. 3x + 2y ≤ 150

(iv) x,y ≥0 (∵ quantity cant be negative)

The feasible region is bounded (ABC)

The corner points of the feasible region are as follows:

Z is minimised at A(15,20) i.e. 15 packets of P and 20 packets of Q should be used to minimise the amount of vitamin A.

The minimum amount of vitamin A is 150 units.

The above information can be expressed with the help of the following table:

Let ‘x’ bags of P and ‘y’ bags of Q be bought.

Cost of P = 250x

Cost of Q = 200y

Cost of mixture = 250x + 200y

Now,

⟹ 3x + 1.5y ≥18

i.e. the minimum requirement of element A from both P and Q combined is 18units, each of which has 3units and 1.5units of element A.

⟹ 2.5x + 11.25y ≥ 45

i.e. the minimum value of element B from both P and Q combined is 45units, each of which contains 2.5units and 11,25units of element B.

⟹ 2x + 3y ≥ 24

i.e. the minimum value of element C from both P and Q combined is 24units, each of which contains 2units and 3 units of element C.

Hence, mathematical formulation of the above LPP is as follows :

Find ‘x’ and ‘y’ that:

Minimises Z = 250x + 200y

Subject to the following constraints:

(i) 3x + 1.5y ≥18

(ii) 2.5x + 11.25y ≥ 45

(iii) 2x + 3y ≥ 24

(iv) x,y ≥0 (∵ quantity cant be negative)

The feasible region is unbounded

The corner points of the feasible region are as follows:

Z is minimised at B(3,6) i.e. 3 bags of P and 6 bags of Q should be purchased to achieve the minimum cost of the mixture per bag.

The minimum cost of the mixture is ₹1950.

The above information can be expressed in the form of the following table:

Let the quantity of X and Y purchased be ‘x’ and ‘y’ kgs

Cost of X = 16x

Cost of Y = 20y

Cost of the mixture = 16x + 20y

Now,

⟹ x + 2y ≥ 10

i.e. the minimum requirement of Vitamin A from the mixture of X and Y is 10 units, each of which contains 1 unit and 2 units of Vitamin A respectively.

⟹ 2x + 2y ≥ 12

i.e. the minimum requirement of Vitamin B from the mixture of X and Y is 12 units, each of which contains 2units of vitamin B each.

⟹ 3x + y ≥ 8

i.e. the minimum requirement of vitamin C from the mixture of X and Y is 8 units, each of which contains 3 units and 1 unit of vitamin C respectively.

Hence, the mathematical formulation of the LPP is as follows :

Find ‘x’ and ‘y’ that:

Minimises Z = 16x + 20y

Subject to the following constraints:

(i) x + 2y ≥ 10

(ii) 2x + 2y ≥ 12

i.e. x + y ≥ 6

(iii) 3x + y ≥ 8

(iv) x,y ≥0 (∵ quantity cant be negative)

The feasible region is unbounded

The corner points of the feasible region is as follows:

Z is smallest at C(2,4)

Let us consider 16x + 20y ≤ 112

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of the mixture is ₹112.

The above information can be expressed with the help of the following table:

Let the number of bags of P and Q chosen be ‘x’ and ‘y’ units.

Nitrogen from P = 3x

Nitrogen from Q = 3.5y

Nitrogen form the mixture = 3x + 3.5y

Now,

⟹ x + 2y ≥ 240

i.e. the minimum requirement of phosphoric acid in the mixture of P and Q is 240 kgs, each of which contains 1kg and 2kgs of phosphoric acid respectively

⟹ 3x + 1.5y ≥ 270

i.e. the minimum requirement of Potash in the mixture of P and Q is 270 kgs, each of which contains 3kgs and 1.5kgs of Potash respectively.

⟹ 1.5x + 2y ≤ 310

i.e. the maximum requirement of Chlorine in the mixture of P and Q is 310 kgs, each of which contains 1.5kgs and 2kgs of Chlorine respectively.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that

Minimises Z = 3x + 3.5y

Subject to the following constraints:

(i) x + 2y ≥ 240

(ii) 3x + 1.5y ≥ 270

(iii) 1.5x + 2y ≤ 310

(iv) x,y ≥0 (∵ quantity cant be negative)

The feasible region is bounded (ABC)

The corner points of the feasible region is as follows:

Z is minimised at B(40,100)

The minimum amount of Nitrogen in the mixture is 470kgs

Let young man drives x km at a speed of 25 km/hr and y km at a speed of 40 km/hr. Clearly,

x, y 0

It is given that, he spends Rs 2 per km if he drives at a speed of 25 km/hr and Rs 5 per km if he drives at a speed of 40 km/hr. Therefore, money spent by him when he travelled x km and y km are Rs 2x and Rs 5y respectively.

It is given that he has a maximum of Rs 100 to spend.

Thus, 2x + 5y 100

Time spent by him when travelling with a speed of 25 km/hr = hr

Time spent by him when travelling with a speed of 40km/hr = hr

Also, the available time is 1 hour.

Or, 40x + 25y1000

The distance covered is Z = x + y which is to be maximized.

Thus, the mathematical formulation of the given linear programming problem is Max Z = x + y subject to

2x + 5y 100

40x + 25y1000

x, y 0

First we will convert inequations as follows:

2x + 5y = 100

40x + 25y = 1000

x = 0 and y = 0.

The region represented by 2x + 5y 100

The line 2x + 5y = 100 meets the coordinate axes at A(50,0) and B(0,20) respectively. By joining these points, we obtain the line 2x + 5y = 100. Clearly (0, 0) satisfies the 2x + 5y = 100. So, the region which contains the origin represents the solution set of the inequation 2x + 5y 100

The region represented by 40x + 25y 1000

The line 40x + 25y = 1000 meets the coordinate axes at C(25,0) and D(0,40) respectively. By joining these points, we obtain the line 2x + y = 12. Clearly (0, 0) satisfies the 40x + 25y = 1000. So, the region which contains the origin represents the solution set of the inequation 40x + 25y 1000

The region represented by x 0, y 0 :

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints

2x + 5y 100, 40x + 25y1000, x 0 and y 0 are as follows

The corner points are O(0,0), B(0,20), E, and C(25,0). The value of Z at these corner points are as follows:

The maximum value of Z is 30 which is attained at E.

Thus, the maximum distance travelled by the young man is 30 kms, if he drives km at a speed of 25 km/hr and km at a speed of 40 km/hr.

Let x units of item A and y units of item B be manufactured. Therefore, x, y 0.

As we are given,

Machines I and II are capable of being operated for at most 12 hours whereas Machine III must operate at least for 5 hours a day.

According to the question, the constraints are

x + 2y 12

2x + y 12

x + y 5

He makes a profit of Rs 6.00 on item A and Rs. 4.00 on item B. Profit made by him in producing x items of A and y items of B is 6x + 4y.

Total profit Z = 6x + 4y which is to be maximized

Thus, the mathematical formulation of the given linear programming problem is

Max Z = 6x + 4y, subject to

x + 2y 12

2x + y 12

x + y 5

x, y 0

First, we will convert the inequations into equations as follows:

x + 2y = 12, 2x + y = 12, x + y = 5, x = 0 and y = 0.

The region represented by x + 2y 12

The line x + 2y = 12 meets the coordinate axes at A(12,0) and B(0,6) respectively. By joining these points, we obtain the line x + y = 12. Clearly (0, 0) satisfies the x + 2y = 12. So, the region which contains the origin represents the solution set of the inequation x + 2y 12

The region represented by 2x + y 12

The line 2x + y = 12 meets the coordinate axes at C(6,0) and D(0,12) respectively. By joining these points, we obtain the line 2x + y = 12. Clearly (0, 0) satisfies the 2x + y = 12. So, the region which contains the origin represents the solution set of the inequation 2x + y 12

The region represented by x + y 5

The line x + y 5 meets the coordinate axes at E(5,0) and F(0,4) respectively. By joining these points, we obtain the line x + y = 5. Clearly (0, 0) satisfies the x + y 5. So, the region which does not contain the origin represents the solution set of the inequation x + y 5

The region represented by x 0, y 0 :

Since every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints

x + 2y 12, 2x + y 12, x + y 5, x, y 0 are as follows.

Thus the maximum profit is of Rs 40 obtained when 4 units each of item A and B are manufactured.

The corner points are D(0,6), I(4,4), C(6,0), G(5,0), and H(0,4). The values of Z at these corner points are as follows:

The maximum value of Z is 40 which is attained at I(4, 4).

Let tailor A work for x days and tailor B work for y days.

In one day, A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants.

Thus in x days, A can stitch 6x shirts and 4x pants whereas in y days B can stitch 10y shirts and 4y pants.

It is given that the minimum requirement of the shirt and pants are respectively 60 and 32.

Thus,

6x + 10y 60

4x + 4y 32

Further it is given that A and B earn Rs 15 and Rs 20 per day respectively. Thus, A earn Rs 15x and B earns Rs 20y.

Let Z denotes the total cost

Z = 15x + 20y

Days cannot be negative.

x,y 0.

MIN Z = 15x + 20y

Subject to

6x + 10y 60

4x + 4y 32

x,y 0

First we will convert inequations into equations as follows:

6x + 10y = 60, 4x + 4y = 32, x = 0, y = 0

Region represented by 6x + 10y 60

The line 6x + 10y = 60 meets the coordinate axes at A(10,0) and B(0,6) respectively. By joining these points we obtain the line 6x + 10y = 60. Clearly (0, 0) satisfies the 6x + 10y 60. So, the region which does not contains the origin represents the solution set of the inequation 6x + 10y 60

Region represented by 4x + 4y 32

The line 4x + 4y = 32 meets the coordinate axes at C(8,0) and D(0,8) respectively. By joining these points we obtain the line 4x + 4y = 32. Clearly (0, 0) satisfies the 4x + 4y 32. So, the region which does not contains the origin represents the solution set of the inequation 4x + 4y 32.

Region represented by x 0, y 0 :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints 6x + 10y 60, 4x + 4y 32

x,y 0 are as follows.

The corner points are D(0,8), E(5,3), A(10,00. The values of Z at these corner points are as follows:

The minimum value of Z is 135 which is attained at E(5,3).

Thus, for minimum labour cost, A should work for 5 days and B should work for 3 days.

Let the factory manufacture x screws of type A and y screws of type B on each day,

Therefore, x 0 and y 0.

The given information can be compiled in a table as follows

4x + 6y 240

6x + 3y 240

The manufacturer can sell a package of screws ‘A’ at a profit of Rs 0.7 and screws ‘B’ at a profit of Re 1.

Total profit, Z = 0.7x + 1y

The mathematical formulation of the given problem is

Maximize Z = 0.7x + 1y

subject to the constraints,

4x + 6y 240

6x + 3y 240

x, y 0

First we will convert the inequations into equations as follows:

4x + 6y = 240, 6x + 3y = 240, x = 0, y = 0.

Region represented by 4x + 6y 240

The line 4x + 6y = 240 meets the coordinate axes at A(60,0) and B(0,40) respectively. By joining these points we obtain the line 4x + 6y = 240. Clearly (0, 0) satisfies the 4x + 6y 240. So, the region which contains the origin represents the solution set of the inequation 4x + 6y 240.

Region represented by 6x + 3y 240

The line 6x + 3y = 240 meets the coordinate axes at C(40,0) and d(0,80) respectively. By joining these points we obtain the line 6x + 3y = 240. Clearly (0, 0) satisfies the 6x + 3y 240. So, the region which contains the origin represents the solution set of the inequation 6x + 3y 240.

Region represented by x 0, y 0 :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints 4x + 6y 240, 6x + 3y 240, x 0,

y 0 are as follows.

The corner points are C(40,0), E(30,20), B(0,40). The values of Z at these corner points are as follows

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws b to get the maximum profit of Rs 410.

Let the company produces x belts of types A and y belts of type B. Number of belts cannot be negative. Therefore, x,y 0.

It is given that leather is sufficient only for 800 belts per day (both A and B combined).

Therefore,

x + y 800

It is given that the rate of production of belts of type B is 1000 per day. Hence the time taken to produce y belts of type B is .

And, since each belt of type A requires twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and therefore, total time taken to produce x belts of type A is

Thus, we have,

Or, 2x + y 1000

Belt A requires fancy buckle and only 400 fancy buckles are available for this per day.

x 400

For Belt of type B only 700 buckles are available per day.

y 700

profits on each type of belt are Rs 2 and Rs 1.50 per belt, respectively. Therefore, profit gained on x belts of type A and y belts of type B is Rs 2x and Rs 1.50y respectively. Hence, the total profit would be Rs(2x + 1.50y). Let Z denote the total profit

Z = 2x + 1.50y

Thus, the mathematical formulation of the given linear programming problem is;

Max Z = 2x + 1.50y subject to

x + y 800

2x + y 1000

x 400

y 700

First we will convert these inequations into equations as follows:

x + y = 800

2x + y = 1000

x = 400

y = 700

Region represented by x + y = 800

The line x + y = 800 meets the coordinate axes at A(800,0) and B(0,800) respectively. By joining these points we obtain the line x + y = 800. Clearly (0, 0) satisfies the x + y 800. So, the region which contains the origin represents the solution set of the inequation x + y 800.

Region represented by 2x + y 1000

The line 2x + y = 1000 meets the coordinate axes at C(500,0) and D(0,1000) respectively. By joining these points we obtain the line 2x + y = 1000. Clearly (0, 0) satisfies the 2x + y 1000. So, the region which contains the origin represents the solution set of the inequation 2x + y 1000.

Region represented by x 400

The line x = 400 will pass through (400,0). The region to the left of the line x = 400 will satisfy the inequation x 400

Region represented by y 700

The line y = 700 will pass through (0,700). The region to the left of the line y = 700

will satisfy the inequation y 700.

Region represented by x 0, y 0 :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x 0 and y 0.

The feasible region determined by the system of constraints x + y 800, 2x + y 1000, x 400,

y 700

The corner points are F(0,700), G(200,600), H(400,200), E(400,0). The values of Z at these corner points are as follows

The maximum value of Z is 1300 which is attained at G(200,600).

Thus, the maximum profit obtained is Rs 1300 when 200 belts of type A and 600 belts of type B are produced.

Let x articles of deluxe model and y articles of an ordinary model be made.

Numbers cannot be negative.

Therefore,

x, y 0

According to the question, the profit on each model of deluxe and ordinary type model are Rs 15 and Rs 10 respectively.

So, profits on x deluxe model and y ordinary models are 15x and 10y.

Let Z be total profit, then,

Z = 15x + 10y

Since, the making of a deluxe and ordinary model requires 2 hrs. and 1 hr work by skilled men, so, x deluxe and y ordinary models require 2x and y hours of skilled men but time available by skilled men is 58 = 40 hours.

So,

2x + y 40 { First Constraint}

Since, the making of a deluxe and ordinary model requires 2 hrs. and 3 hrs work by semi skilled men, so, x deluxe and y ordinary models require 2x and 3y hours of skilled men but time available by skilled men is 108 = 80 hours.

So,

2x + 3y 80 {Second constraint}

Hence the mathematical formulation of LPP is,

Max Z = 15x + 10y

subject to constraints,

2x + y 40

2x + 3y 80

x, y 0

Region 2x + y 40: line 2x + 4y = 40 meets axes at (20,0), (0,40) respectively. Region containing origin represents 2x + 3y 40 as (0,0) satisfies 2x + y 40

Region 2x + 3y 80: line 2x + 3y = 80 meets axes at (40,0), (0,) respectively. Region containing origin represents 2x + 3y 80.

The corner points are (20,0), P(10,20), (0,).

The value of Z = 15x + 10y at these corner points are

The maximum value of Z is 300 which is attained at P(10,20).

Thus, maximum profit is obtained when 10 units of deluxe model and 20 units of ordinary model is produced.

Let the required number of tea cups of Type A and B are x and y respectively.

Since, the profit on each cup A is 75 paise and that on each cup B is 50 paise. So, the profit on x tea cup of type A and y tea cup of type B are 75x and 50y respectively.

Let total profit on tea cups be Z, so

Z = 75x + 50y

Since, each tea cup of type A and B require to work machine I for 12 and 6 minutes respectively so, x tea cups of Type A and y tea cups of Type B require to work on machine I for 12x and 6y minutes respectively.

Total time available on machine I is 660 = 360 minutes. So,

12x + 6y 360 {First Constraint}

Since, each tea cup of type A and B require to work machine II for 18 and 0 minutes respectively so, x tea cups of Type A and y tea cups of Type B require to work on machine IIII for 18x and 0y minutes respectively.

Total time available on machine I is 660 = 360 minutes. So,

18x + 0y 360

x 20 {Second Constraint}

Since, each tea cup of type A and B require to work machine III for 6 and 9 minutes respectively so, x tea cups of Type A and y tea cups of Type B require to work on machine I for 6x and 9y minutes respectively.

Total time available on machine I is 660 = 360 minutes. So,

6x + 9y 360 {Third Constraint}

Hence mathematical formulation of LPP is,

Max Z = 75x + 50y

subject to constraints,

12x + 6y 360

x 20

6x + 9y 360

x,y 0 [Since production of tea cups can not be less than zero]

Region 12x + 6y 360: line 12x + 6y = 360 meets axes at A(30,0), B(0,60) respectively. Region containing origin represents 12x + 6y 360 as (0,0) satisfies 12x + 6y 360

Region x 20: line x = 20 is parallel to y axis and meets x - axes at C(20,0). Region containing origin represents x 20

as (0,0) satisfies x 20.

Region 6x + 9y 360: line 6x + 9y = 360 meets axes at E(60,0), F(0,40) respectively. Region containing origin represents 6x + 9y 360 as (0,0) satisfies 6x + 9y 360.

Region x,y 0: it represents the first quadrant.

The shaded region is the feasible region determined by the constraints,

12x + 6y 360

x 20

6x + 9y 360

x,y 0

The corner points are F(0,40), G(15,30), H(20,20), C(20,0).

The values of Z at these corner points are as follows

Here Z is maximum at G(15,30).

Therefore, 15 teacups of Type A and 30 tea cups of Type B are needed to maximize the profit.

Let required number of machine A and B are x and y respectively.

Since, products of each machine A and B are 60 and 40 units daily respectively. So, production by by x number of machine A and y number of machine B are 60x and 40y respectively.

Let Z denotes total output daily, so,

Z = 60x + 40y

Since, each machine of type A and B requires 1000sq. m and 1200 s. m area so, x machine of type A and y machine of type B require 1000x and 1200y sq. m area but,

Total available area for machine is 7600 sq. m. So,

1000x + 1200y 7600

or, 5x + 6y 38. {First Constraint}

Since each machine of type A and B requires 12 men and 8 men to work respectively. So, x machine of type A and y machine of type B require 12x and 8y men to work respectively.

But total men available for work are 72.

So,

12x + 8y 72

3x + 2y 18 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 50x + 40y

Subject to constraints,

5x + 6y 38

3x + 2y 18

x,y 0 [Since number of machines can not be less than zero]

Region 5x + 6y 38: line 5x + 6y = 38 meets the axes at A(,0), B(0,) respectively.

Region containing the origin represents 5x + 6y 38 as origin satisfies 5x + 6y 38

Region 3x + 2y 18: line 3x + 2y = 18 meets the axes at C(6,0), D(0,9) respectively.

Region containing the origin represents 3x + 2y 18 as origin satisfies 3x + 2y 18.

Region x,y 0: it represents the first quadrant.

Shaded region represents the feasible region.

The corner points are O(0,0), B(0,), E(4,3), C(6,0).

Thus the values of Z at these corner points are as follows:

The maximum value of Z is 360 which is attained at E(4,3), C(6,0).

Thus,the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A and 0 units of type B are manufactured.

Let required number of goods A and B are x and y respectively.

Since, profits of each A and B are Rs. 40 and Rs. 50 respectively. So, profits on x number of type A and y number of type B are 40x and 50y respectively.

Let Z denotes total output daily, so,

Z = 40x + 50y

Since, each A and B requires 3 grams and 1 gram of silver respectively. So, x of type A and y of type B require 3x and y of silver respectively. But,

Total silver available is 9 grams. So,

3x + y 9 {First Constraint}

Since each A and B requires 1 gram and 2 grams of gold respectively. So, x of type A and y of type B require x and 2y respectively.

But total gold available is 8 grams.

So,

x + 2y 8 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 40x + 50y

Subject to constraints,

3x + y 9

x + 2y 8

x,y 0 [Since production of A and B can not be less than zero]

Region 3x + y 9: line 3x + y = 9 meets the axes at A(3,0), B(0,9) respectively.

Region containing the origin represents 3x + y 9

as origin satisfies 3x + y 9.

Region x + 2y 8: line x + 2y = 8 meets the axes at C(8,0), D(0,4) respectively.

Region containing the origin represents x + 2y 8 as origin satisfies x + 2y 8.

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), D(0,4), E(2,3), A(3,0)

The values of Z at these corner points are as follows

The maximum value of Z is 230 which is attained at E(2,3).

Thus the maximum profit is of Rs 230 when 2 units of Type A 3 uniits of Type B are produced.

Let daily production of chairs and tables be x and y respectively.

Since, profits of each chair and table is Rs. 3 and Rs. 5 respectively. So, profits on x number of type A and y number of type B are 3x and 5y respectively.

Let Z denotes total output daily, so,

Z = 3x + 5y

Since, each chair and table requires 2 hrs and 3 hrs on machine A respectively. So, x number of chair and y number of table require 2x and 4y hrs on machine A respectively. But,

Total time available on Machine A is 16 hours. So,

2x + 3y 16

x + 2y 8 {First Constraint}

Since, each chair and table requires 6 hrs and 2 hrs on machine B respectively. So, x number of chair and y number of table require 6x and 2y hrs on machine B respectively. But,

Total time available on Machine B is 30 hours. So,

6x + 2y 30

3x + y 15 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 3x + 5y

Subject to constraints,

x + 2y 8

3x + y 15

x,y 0 [Since production of chairs and tables can not be less than zero]

Region x + 2y 8: line x + 2y = 8 meets the axes at A(8,0), B(0,4) respectively.

Region containing the origin represents x + 2y 8

as origin satisfies x + 2y 8.

Region 3x + y 15: line 3x + y = 15 meets the axes at C(5,0), D(0,15) respectively.

Region containing the origin represents 3x + y 15 as origin satisfies 3x + y 15

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), B(0,4), E(), and C(5,0).

The values of Z at these corner points are as follows,

The maximum value of Z is 22.2 which is attained at E().

Thus the maximum profit of Rs 22.2 when units of chair and units of table are produced.

Let required production of chairs and tables be x and y respectively.

Since, profits of each chair and table is Rs. 45 and Rs. 80 respectively. So, profits on x number of type A and y number of type B are 45x and 80y respectively.

Let Z denotes total output daily, so,

Z = 45x + 80y

Since, each chair and table requires 5 sq. ft and 80 sq. ft of wood respectively. So, x number of chair and y number of table require 5x and 80y sq. ft of wood respectively. But,

But 400 sq. ft of wood is available. So,

5x + 80y 400

x + 4y 80 {First Constraint}

Since, each chair and table requires 10 and 25 men - hours respectively. So, x number of chair and y number of table require 10x and 25y men - hours respectively. But, only 450 hours are available . So,

10x + 25y 450

2x + 5y 90 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 45x + 80y

Subject to constraints,

x + 4y 80

2x + 5y 90

x,y 0 [Since production of chairs and tables can not be less than zero]

Region x + 4y 80: line x + 4y = 80 meets the axes at A(80,0), B(0,20) respectively.

Region containing the origin represents x + 4y 80 as origin satisfies x + 4y 80

Region 2x + 5y 90: line 2x + 5y = 90 meets the axes at C(45,0), D(0,20) respectively.

Region containing the origin represents 2x + 5y 90

as origin satisfies 2x + 5y 90

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), D(0,18), C(45,0).

The values of Z at these corner points are as follows:

The maximum value of Z is 2025 which is attained at C(45,0).

Thus maximum profit of Rs 2025 is obtained when 45 units of chairs and no units of tables are produced.

Let required production of product A and B be x and y respectively.

Since profit on each product A and B are Rs. 3 and Rs. 4 respectively. So, profits on x number of type A and y number of type B are 3x and 4y respectively.

Let Z denotes total output daily, so,

Z = 3x + 4y

Since, each A and B requires 4 minutes each on machine . So, x of type A and y of type B require 4x and 4y minutes respectively. But,

Total time available on machine is 8 hours 20 minutes = 500 minutes.

So,

4x + 4y 500

x + y 125 {First Constraint}

Since, each A and B requires 8 minutes and 4 minutes on machine respectively. So, x of type A and y of type B require 8x and 4y minutes respectively. But,

Total time available on machine is 10 hours = 600 minutes.

So,

8x + 4y 600

2x + y 150 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 3x + 4y

Subject to constraints,

x + y 125

2x + y 150

x,y 0 [Since production of A and B can not be less than zero]

Region x + y 125: line x + y = 125 meets the axes at A(125,0), B(0,125) respectively.

Region containing the origin represents x + y 125 as origin satisfies x + y 125.

Region 2x + y 150: line 2x + y = 150 meets the axes at C(75,0), D(0,150) respectively.

Region containing the origin represents 2x + y 150 as origin satisfies 2x + y 150.

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), B(0,125), E(25,100), and C(75,0).

The vaues of Z at these corner points are as follows:

The maximum value of Z is 500 which is attained at B(0,125).

Thus, the maximum profit is Rs 500 obtained when no units of product A and 125 units of product B are manufactured.

Let x units of item A and y units of item B were manufactured.

Numbers of items cannot be negative. Therefore,

x, y 0

The given information can be tabulated as follows:

Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.

Therefore, the constraints are

3x + 2y 210

4x + 4y 300

y 65

A and B can make profit of Rs 20 and Rs 30 per unit respectively.

Therefore, profit gained from x units of item A and y units of item B is Rs 20x and 30y respectively.

Total Profit = Z = 20x + 30y which according to question is to be maximised.

Thus the mathematical formulation of the given LPP is,

Max Z = 20x + 30y

Subject to constraints

3x + 2y 210

4x + 4y 300

y 65

x, y 0

Region represented by 3x + 2y 210: The line 3x + 2y = 210 meets the axes at A(70,0), B(0,105) respectively.

Region containing the origin represents 3x + 2y 210 as origin satisfies 3x + 2y 210.

Region represented by 4x + 4y 300: The line 4x + 4y = 300 meets the axes at C(75,0), D(0,75) respectively.

Region containing the origin represents 4x + 4y 300 as origin satisfies 4x + 4y 300

y = 65 is the line passing through the point E(0,65) and is parallel to X - axis.

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), E(0,65), G(10,65), F(60,15) and A(70,0).

The values of Z at these corner points are as follows:

The maximum value of Z is 2150 which is attained at G(10,65).

Thus, the maximum profit is Rs. 2150 obtained when 10 units of item A and 65 units of item B are manufactured.

Let number of product I and product II are x and y respectively.

Since, profits on each product I and II are 2 and 3 monetary unit. So, profits on x number of Product I and y number of Product II are 2x and 3y respectively.

Let Z denotes total output daily, so,

Z = 2x + 3y

Since, each I and II requires 2 and 4 units of resources A. So, x units of product I and y units of product II requires 2x and 4y minutes respectively. But, maximum available quantity of resources A is 20 units.

So,

2x + 4y 20

x + 2y 10 {First Constraint}

Since, each I and II requires 2 and 2 units of resources B. So, x units of product I and y units of product II requires 2x and 2y minutes respectively. But, maximum available quantity of resources A is 12 units.

So,

2x + 2y 12

x + y 6 {Second Constraint}

Since, each units of product I requires 4 units of resources C. It is not required by product II. So, x units of product I require 4x units of resource C. But, maximum available quantity of resources C is 16 units.

So,

4x 16

x 4 {Third Constraint}

Hence mathematical formulation of LPP is,

Max Z = 2x + 3y

Subject to constraints,

x + 2y 10

x + y 6

x 4

x, y 0 [ Since production for I and II can not be less than zero]

Region represented by x + 2y 10: The line x + 2y = 10 meets the axes at A(10,0), B(0,5) respectively.

Region containing the origin represents x + 2y 10 as origin satisfies x + 2y 10.

Region represented by x + y 6: The line x + y = 6 meets the axes at C(6,0), D(0,6) respectively. Region containing the origin represents x + y 6 as origin satisfies x + y 6

Region x,y 0: it represents the first quadrant.

The corner points are O(0,0), B(0,5), G(2,4), F(4,2), and E(4,0).

The values of Z at these corner points are as follows:

The maximum value of Z is 16 which is attained at G (12,4).

Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufactured.

Let the sale of hand cover edition be ‘h’ and that of paperback editions be ‘t’.

SP of a hard cover edition of the textbook = Rs 72

SP of a paperback edition of the textbook = Rs 40

Cost to the publisher for hard cover edition = Rs 56

Cost to the publisher for a paperback edition = Rs 28

Weekly cost to the publisher = Rs 9600

Profit to be maximized, Z = (72 - 56)h + (40 - 28)t – 9600

Z = 16h + 12t - 9600

5(h + t) 4800

10h + 2t 4800.

The corner Points are O(0,0), (0,960), (360,600) and (480,00).

The values of Z at these corner points are as follows:

The maximum value of Z is 3360 which is attained at (360,600).

The maximum profit is 3360 which is obtained by selling 360 copies of hardcover edition and 600 copies paperback edition.

The above LPP can be presented in a table below,

Hence mathematical formulation of LPP is,

Max Z = x + y

Subject to constraints,

2x + y 12

5x + 8y 7.4

x + 66y 24

x, y 0 [ Since production can not be less than zero]

The corner points are B(0,12), P(5.86, 0.27), E(24,0)

The values of Z at these corner points are as follows:

The minimum value of Z is 6.13 but the region is unbounded so check whether x + y 6.13

Clearly, it can be seen that it does not has any common region.

So, x = 5.86, y = 0.27

This is the least quantity of pill A and B.

Codeine quantity = x + 66y = 5.86 + (660.27) = 24(approx).

Let required quantity of compound A and B are x and y kg.

Since, cost of one kg of compound A and B are Rs 4 and Rs 6 per kg. So,

Cost of x kg of compound A and y kg of compound B are Rs 4x and Rs 6

respectively.

Let Z be the total cost of compounds, so,

Z = 4x + 6y

Since, compound A and B contain 1 and 2 units of ingredient C per kg

respectively, So x kg of compound A and y kg of compound B contain x and 2y

units of ingredient C respectively but minimum requirement of ingredient C is 80

units, so,

x + 2y 80 {first constraint}

Since, compound A and B contain 3 and 1 units of ingredient D per kg

respectively, So x kg of compound A and y kg of compound B contain 3x and y

units of ingredient D respectively but minimum requirement of ingredient C is 75

units, so,

3x + y 75 {second constraint}

Hence, mathematical formulation of LPP is,

Min Z = 4x + 6y

Subject to constraints,

x + 2y 80

3x + y 75

x, y 0 [Since production can not be less than zero]

Region x + 2y 80: line x + 2y = 80 meets axes at A(80,0), B(0,40)

respectively. Region not containing origin represents x + 2y 80 as (0,0) does

not satisfy x + 2y 80.

Region 3x + y 75: line 3x + y = 75 meets axes at C(25,0), D(0,75)

respectively. Region not containing origin represents 3x + y 75 as (0,0) does

not satisfy 3x + y 75.

Region x,y 0: it represents first quadrant.

The corner points are D(0,75), E(14,33), A(80,0).

The values at Z at these corner points are as follows:

Corner Point Z = 4x + 6y

D 450

E 254

A 320

The minimum value of Z is 254 which is attained at E(14,33).

Thus, the minimum cost is Rs 254 obtained when 14 units of compound A

and 33 units compound B are produced.

Let the company manufacture x souvenirs of Type A and y souvenirs of Type B.

Therefore, x 0, y 0

The given information can be compiled in a table as follows:

The profit on Type A souvenirs is 50 paisa and on Type B souvenirs is 60 paisa. Therefore, profit gained on x souvenirs of Type A and y

souvenirs of Type B is Rs 0.50x and Rs 0.60y respectively.

Total Profit, Z = 0.5x + 0.6y

The mathematical formulation of the given problem is,

Max Z = 0.5x + 0.6y

Subject to constraints,

5x + 8y 200

10x + 8y 240

x 0, y 0

Region 5x + 8y 200: line 5x + 8y = 200 meets axes at A(40,0), B(0,25) respectively. Region containing origin represents the solution of the inequation 5x + 8y 200 as (0,0) satisfies 5x + 8y 200.

Region 10x + 8y 240: line 10x + 8y = 240 meets axes at C(24,0), D(0,30) respectively. Region containing origin represents the solution of the inequation 10x + 8y 240 as (0,0) satisfies 10x + 8y 240.

Region x,y 0: it represents first quadrant.

The corner points of the feasible region are O(0,0), B(0,25), E(8,20), C(24,0).

The values of Z at these corner points are as follows:

The maximum value of Z is attained at E(8,20).

Thus, 8 souvenirs of Type A and 20 souvenirs of Type B should be produced each day to get the maximum profit of Rs 16.

Let x units of product A and y units of product B were manufactured.

Number of units cannot be negative.

Therefore, x,y 0.

According to question, the given information can be tabulated as:

Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.

Further, it is given that there is a permanent order for 14 units of Product A and 16 units of product B.

Therefore, the constraints are,

200x + 300y 10000,

0.5x + y 40

x 14

y 16.

If the profit on each of product A is Rs 20 and on product B is Rs 30. Therefore, profit gained on x units of product A and y units of product B is Rs 20x and Rs 30y respectively.

Total profit = 20x + 30y which is to be maximized.

Thus, the mathematical formulation of the given LPP is,

Max Z = 20x + 30y

Subject to constraints,

200x + 300y 10000,

0.5x + y 40

x 14

y 16

x,y 0.

Region 200x + 300y 10000: line 200x + 300y = 10000 meets the axes at A(50,0), B(0,) respectively.

Region not containing origin represents 200x + 300y 10000 as (0,0) does not satisfy 200x + 300y 10000.

Region 0.5x + y 40: line 0.5x + y = 40 meets the axes at C(80,0), D(0,40) respectively.

Region containing origin represents 0.5x + y 40 as (0,0) satisfies 0.5x + y 40.

Region represented by x 14,

x = 14 is the line passes through (14,0) and is parallel to the Y - axis. The region to the right of the line x = 14 will satisfy the inequation.

Region represented by y 16,

y = 14 is the line passes through (16,0) and is parallel to the X - axis. The region to the right of the line y = 14 will satisfy the inequation.

Region x,y 0: it represents first quadrant.

The corner points of the feasible region are E(26,16), F(48,16), G(14,33), H(14,24).

The values of Z at these corner points are as follow:

The maximum value of Z is Rs 1440 which is attained at F(48,16).

Thus, the maximum profit is Rs 1440 obtained when 48 units of product A and 16 units of product B are manufactured.

Let x trunks of first type and y trunks of second type were manufactured. Number of trunks cannot be negative.

Therefore, x, y 0

According to the question, the given information can be tabulated as

Therefore, the constraints are,

3x + 3y 18

3x + 2y 15.

He earns a profit of Rs 30 and Rs 25 per trunk of the first type and the second type respectively. Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs 30x and Rs 25y respectively.

Total profit Z = 30x + 25y which is to be maximized.

Thus, the mathematical formulation of the given LPP is

Max Z = 30x + 25y

Subject to

3x + 3y 18

3x + 2y 15

x, y 0

Region 3x + 3y 18: line 3x + 3y = 18 meets axes at A(6,0), B(0,6) respectively. Region containing origin represents the solution of the inequation 3x + 3y 18 as (0,0) satisfies 3x + 3y 18.

Region 3x + 2y 15: line 3x + 2y = 15 meets axes at C(5,0), D(0,) respectively. Region containing origin represents the solution of the inequation 3x + 2y 15 as (0,0) satisfies 3x + 2y 15.

Region x,y 0: it represents first quadrant.

The corner points are O(0,0), B(0,6), E(3,3), and C(5,0).

The values of Z at these corner points are as follows:

The maximum value of Z is 165 which is attained at E(3,3).

Thus, the maximum profit is of Rs 165 obtained when 3 units of each type of trunk is manufactured.

Let production of each bottle of A and B are x and y respectively.

Since profits on each bottle of A and B are Rs 8 and Rs 7 per bottle respectively. So, profit on x bottles of A and y bottles of of B are 8x and 7y respectively. Let Z be total profit on bottles so,

Z = 8x + 7y

Since, it takes 3 hours and 1 hour to prepare enough material to fill 1000 bottles of Type A and Type B respectively, so x bottles of A and y bottles of B are preparing is hours and hours respectively, bout only 66 hours are available, so,

3x + y 66000

Since raw materials available to make 2000 bottles of A and 4000 bottles of B but there are 45000 bottles in which either of these medicines can be put so,

x 20000

y 40000

x + y 45000

x,y 0. [ Since production of bottles can not be negative]

Hence mathematical formulation of the given LPP is,

Max Z = 8x + 7y

Subject to constraints,

3x + y 66000

x 20000

y 40000

x + y 45000

x,y 0

Region 3x + y66000: line 3x + y = 66000 meets the axes at A(22000,0), B(0,66000) respectively.

Region containing origin represents 3x + y 10000 as (0,0) satisfy 3x + y 66000

Region x + y 45000: line x + y = 45000 meets the axes at C(45000,0), D(0,45000) respectively.

Region towards the origin will satisfy the inequation as (0,00 satisfies the inequation

Region represented by x 20000,

x = 20000 is the line passes through (20000, 0) and is parallel to the Y - axis. The region towards the origin will satisfy the inequation.

Region represented by y 40000,

y = 40000 is the line passes through (0,40000) and is parallel to the X - axis. The region towards the origin will satisfy the inequation.

Region x,y 0: it represents first quadrant.

The corner points are O(0,0), B(0,40000), G(10500,34500), H(20000,6000), A(20000,0).

The values of Z at these corner points are,

The maximum value of Z is 325500 which is attained at G(10500, 34500).

Thus the maximum profit is Rs 325500 obtained when 10500 bottles of A and 34500 bottles of B are manufactured.

Let required number of first class and economy class tickets be x and y respectively.

Each ticket of first class and economy class make profit of Rs 400 and Rs 600 respectively.

So, x ticket of first class and y tickets of economy class make profit of Rs 400x and Rs 600y respectively.

Let total profit be Z = 400x + 600y

Given, aeroplane can carry a minimum of 200 passengers, so,

x + y 200

Given airline reserves at least 20 seats for first class, so,

x 20

Also, at least 4 times as many passengers prefer to travel by economy class to the first class, so

y 4x

Hence the mathematical formulation of the LPP is

Max Z = 400x + 600y

Subject to constraints

x + y 200

y 4x

x 20

x,y 0 [ since seats in both the classes can not be zero]

Region represented by x + y 200: the line x + y = 200 meets the axes at A(200,0), B(0,200). Region containing origin represents x + y 200 as (0,0) satisfies x + y 200.

Region represented by x 20: line x = 20 passes through (20,0) and is parallel to y axis. The region to the right of the line x = 20 will satisfy the inequation x 20

Region represented by y 4x: line y = 4x passes through (0,0). The region above the line y = 4x will satisfy the inequation y 4x

Region x,y 0: it represents the first quadrant.

The corner points are C(20,80), D(40,160), E(20,180).

The values of Z at these corner points are as follows:

The maximum value of Z is attained at E(20, 180).

Thus, the maximum profit is Rs 116000 obtained when 20 first class tickets and 180 economy class tickets are sold.

Let x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.

The quantity of fertilizers can not be negative.

So, x,y 0

A gardener has a supply of fertilizer of type I which consists of 10% nitrogen and Type II consists of 5% nitrogen, and he needs at least 14 kg of nitrogen for his crop.

So,

(10100) + (5100) 14

Or, 10x + 5y 1400

A gardener has a supply of fertilizer of type I which consists of 6% phosphoric acid and Type II consists of 10% phosphoric acid, and he needs at least 14 kg of phosphoric acid for his crop.

So,

(6100) + (10100) 14

Or, 6x + 10y 1400

Therefore, A/Q, constraints are,

10x + 5y 1400

6x + 10y 1400

If the Type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. Therefore, the cost of x kg of Type I fertilizer and y kg of Type II fertilizer is Rs0.60x and Rs 0.40y respectively.

Total cost = Z(let) = 0.6x + 0.4y is to be minimized.

Thus the mathematical formulation of the given LPP is,

Min Z = 0.6x + 0.4y

Subject to the constraints,

10x + 5y 1400

6x + 10y 1400

x,y 0

The region represented by 6x + 10y 1400: line 6x + 10y = 1400 passes through A(,0) and B(0,140). The region which doesn’t contain the origin represents the solution of the inequation 6x + 10y 1400

As (0,0) doesn’t satisfy the inequation 6x + 10y 1400

Region represented by 10x + 5y 1400: line 10x + 5y = 1400 passes through C(140,0) and D(0,280). The region which doesn’t contain the origin represents the solution of the inequation 10x + 5y 1400

As (0,0) doesn’t satisfy the inequation 10x + 5y 1400

The region, x,y 0: represents the first quadrant.

The corner points are D(0,280), E(100,80), A(,0)

The values of Z at these points are as follows:

The minimum value of Z is Rs 92 which is attained at E(100,80)

Thus, the minimum cost is Rs92 obtained when 100 kg of Type I fertilizer and 80 kg of TypeII fertilizer is supplied.

Let Anil invests Rs x and Rs y in saving certificate (SC) and National saving bond (NSB) respectively.

Since, the rate of interest on SC is 8% annual and on NSB is 10% annual. So, interest on Rs x of SC is and Rs y of NSB is per annum.

Let Z be total interest earned so,

Z = +

Given he wants to invest Rs 12000 is total

x + y 12000

According to the rules he has to invest at least Rs 2000 in SC and at least Rs 4000 in NSB.

x 2000

y 4000

Hence the mathematical formulation of LPP is to find x and y which

Maximizes Z

Max Z = +

Subject to constraints

x 2000

y 4000

x + y 12000

x,y 0

The region represented by x 2000: line x = 2000 is parallel to the y - axis and passes through (2000,0).

The region not containing the origin represents x 2000

As (0,0) doesn’t satisfy the inequation x 2000

The region represented by y 4000: line y = 4000 is parallel to the x - axis and passes through (0,4000).

The region not containing the origin represents y 4000

As (0,0) doesn’t satisfy the inequation y 4000

Region represented by x + y 12000: line x + y = 12000 meets axes at A(12000,0) and B(0,12000) respectively. The region which contains the origin represents the solution set of x + y 12000

as (0,0) satisfies the inequality x + y 12000.

Region x,y 0 is represented by the first quadrant.

The corner points are E(2000,10000), C(2000,4000), D(8000,4000).

The values of Z at these corner points are as follows:

The maximum value of Z is Rs 1160 which is attained at E(2000,10000).

Thus the maximum earning is Rs1160 obtained when Rs 2000 were invested in SC and Rs 10000 in NSB.

Let the required number of trees of Type A and B be Rs x and Rs y respectively.

Number of trees cannot be negative.

x,y 0.

To plant tree of Type A requires 10 sq. m and Type B requires 20 sq. m of ground per tree. And it is given that a man owns a field of area 1000 sq. m. Therefore,

10x + 20y 1000

x + 2y100

Type A costs Rs 20 per tree and Type B costs Rs 25 per tree. Therefore, x trees of type A and y trees of type B cost Rs 20x and Rs 25y respectively. A man has a sum of Rs 1400 to purchase young trees.

20x + 25y 1400

4x + 5y280

Thus the mathematical formulation of the given LPP is

Max Z = 40x - 20x + 60y - 25y = 20x + 35y

Subject to,

x + 2y100

4x + 5y280

x,y 0

Region 4x + 5y280: line 4x + 5y280 meets axes at (70,0), (0,56) respectively.

The region containing origin represents 4x + 5y280 as (0,0) satisfies 4x + 5y280.

Region x + 2y100: line x + 2y = 100 meets axes at (100,0), (0,50) respectively.

Region containing origin represents x + 2y100 as (0,0) satisfies x + 2y100

Region x,y 0: it represents the first quadrant.

The corner points are (70,0), P(20,40), (0,50)

The values of Z at these corner points are as follows:

The maximum value of Z is 1800 which is attained at P(20,40).

Thus the maximum profit is Rs 1800 obtained when Rs 20 were involved in Type A and Rs 40 were involved in Type II.

Let godown A supply x and y quintals of grain to shop D and E respectively. Remaining grain storage of A is 100 – x – y. So, now it is supplied to shop F by godown A.

Now, remaining requirement of shop D is 60 – x which is supplied by godown B.

Remaining requirement of shop E is 50 – y which is supplied by godown B.

Remaining requirement of shop F is 40 – (100 – x – y)

= x + y - 60 which is supplied by godown B.

where,

x ≥ 0, y ≥ 0 and 100 – x – y ≥ 0

⇒ x ≥ 0, y ≥ 0 and x + y ≤ 100

60 – x ≥ 0, 50 – y ≥ 0 and x + y – 60 ≥ 0

⇒x ≤ 60, y ≤ 50 and x + y ≥ 60

This can be illustrated by this:

Cost = Number of quintals * Cost of transportation per quintal

Total transportation cost z is given by,

z = 6x + 3y + 2.5(100 – x – y) + 4(60 – x) + 2(50 – y) + 3(x + y – 60)

⇒ z = 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180

⇒ z = 2.5x + 1.5y + 410

We need to minimize the cost

Hence, mathematical formulation of LPP is

Minimize z = 2.5x + 1.5y + 410

subject to the constraints,

x + y ≥ 60

y ≤ 50

x ≤ 60

x + y ≤ 100

x, y≥0

The feasible region determined by the system of constraints is as follows:

The corner points of enclosed region are A(60, 0) , B(60, 40), C(50, 50) and D(10,50)

The value of z at these corners points is as follows:

Case 1: A(60, 0)

z = 2.5x + 1.5y + 410

⇒ z = 2.5(60) + 1.5(0) + 410

⇒ z = 150 + 0 + 410

⇒ z = 560

Case 2: B(60, 40)

z = 2.5x + 1.5y + 410

⇒ z = 2.5(60) + 1.5(40) + 410

⇒ z = 150 + 60 + 410

⇒ z = 620

Case 3: C(50, 50)

z = 2.5x + 1.5y + 410

⇒ z = 2.5(50) + 1.5(50) + 410

⇒ z = 125 + 75 + 410

⇒ z = 610

Case 4: D(10, 50)

z = 2.5x + 1.5y + 410

⇒ z = 2.5(10) + 1.5(50) + 410

⇒ z = 25 + 75 + 410

⇒ z = 510

The value of z is minimum in fourth case at point D(10, 50)

As, x = 10, y = 50

Godown A supplies to:

Shop D = x = 10 quintals

Shop E = y = 50 quintals

Shop F = 100 – x – y = 100 – 10 – 50 = 40 quintals

Godown B supplies to:

Shop D = 60 – x = 60 – 10 = 50 quintals

Shop E = 50 – y = 50 – 50 = 0 quintals

Shop F = x + y – 60 = 10 + 50 – 60 = 0 quintals

Minimum cost for transportation of these quintals to their respective shops = Rs. 510

Let factory A supply x and y packets of medicines to agency P and Q respectively. Remaining packets of factory A are 60 – x – y. So, now they will be supplied to agency R by factory A.

Now, remaining packets requirement of agency P is 40 – x which will be supplied by factory B.

Remaining packets requirement of agency Q is 40 – y which will be supplied by factory B.

Remaining packets requirement of agency R is 50 – (60 – x – y)

= x + y – 10 which will be supplied by factory B.

where,

x ≥ 0, y ≥ 0 and 60 – x – y ≥ 0

⇒ x ≥ 0, y ≥ 0 and x + y ≤ 60

40 – x ≥ 0, 40 – y ≥ 0 and x + y – 10 ≥ 0

⇒x ≤ 40, y ≤ 40 and x + y ≥ 10

This can be illustrated by this:

Cost = Number of packets * Cost of transportation per packet

Total transportation cost z is given by,

z = 5x + 4y + 3(60 – x – y) + 4(40 – x) + 2(40 – y) + 5(x + y – 10)

⇒ z = 5x + 4y + 180 – 3x – 3y + 160 – 4x + 80 – 2y + 5x + 5y – 50

⇒ z = 3x + 4y + 370

We need to minimize the cost

Hence, mathematical formulation of LPP is

Minimize z = 3x + 4y + 370

subject to the constraints,

x+y≥10

y≤40

x≤40

x+y≤60

x,y≥0

The feasible region determined by the system of constraints is as follows:

The corner points of the enclosed region are A(10, 0) , B(40, 0), C(40, 20), D(20,40), E(0, 40) and F(0, 10)

The value of z at these corners points is as follows:

Case 1: A(10, 0)

z = 3x + 4y + 370

⇒ z = 3(10) + 4(0) + 370

⇒ z = 30 + 0 + 370

⇒ z = 400

Case 2: B(40, 0)

z = 3x + 4y + 370

⇒ z = 3(40) + 4(0) + 370

⇒ z = 120 + 0 + 370

⇒ z = 490

Case 3: C(40, 20)

z = 3x + 4y + 370

⇒ z = 3(40) + 4(20) + 370

⇒ z = 120 + 80 + 370

⇒ z = 570

Case 4: D(20, 40)

z = 3x + 4y + 370

⇒ z = 3(20) + 4(40) + 370

⇒ z = 60 + 160 + 370

⇒ z = 590

Case 5: E(0, 40)

z = 3x + 4y + 370

⇒ z = 3(0) + 4(40) + 370

⇒ z = 0 + 160 + 370

⇒ z = 530

Case 4: F(0, 10)

z = 3x + 4y + 370

⇒ z = 3(0) + 4(10) + 370

⇒ z = 0 + 40 + 370

⇒ z = 410

The value of z is minimum in first case at point A(10, 0)

As, x = 10, y = 0

Factory A supplies to:

Agency P = x = 10 packets

Agency Q = y = 0 packets

Agency R = 60 – x – y = 60 – 10 – 0 = 50 packets

Godown B supplies to:

Agency P = 40 – x = 40 – 10 = 30 packets

Agency Q = 40 – y = 40 – 0 = 40 packets

Agency R = x + y – 10 = 10 + 0 – 10 = 0 packets

Minimum cost for transportation of these packets to their respective agencies = Rs. 400

Given inequation is 2x + y > 5.

Now, we convert the inequation into an intercept line equation form, we can clearly see the intercepts of the inequation on x-axis and y-axis.

2x + y > 5

[dividing the whole inequation by 5]

Therefore, from the above inequation, we can say that, 2.5 and 5 are the intercepts of the x-axis and y-axis respectively.

Now by plotting these on the graph, we can clearly see the graph of the inequation.

From the graph, it is clear that, the solution set of the inequation

2x + y > 5 is the open half plane not containing origin i.e. option B.

Given,

To define the objective of a Linear programming Problem.

As per the definition of the Linear Programming Problem,

A Linear programming problem is a linear function (also known an objective function) subjected to certain constraints for which we need to find an optimal solution ( i.e. either a maximum/minimum value) depending on the requirement of the problem.

From the above definition, we can clearly say that, Linear programming problem’s objective is to either maximize/ minimize a given objective function, which means to optimize a function to get an optimum solution.

Hence the answer is option B.

Given sets are

• {(x,y) : x² + y²≥ 1}

• {(x, y) :y² ≥ x}

• {(x, y) : 3x² + 4y² ≥ 5}

• {(x, y) : y ≥ 2, y ≤ 4}

A convex set, is nothing but whose solution set is in the shape of a convex polygon.

If we map these functions on a graph, we can clearly find the set with a convex solution set.

• f : {(x,y) : x² + y²≥ 1}

From the graph, it is evident that the solution set which is the shaded region is not convex.

• g:{(x, y) :y² ≥ x}

From the graph, it is evident that the solution set which is the blue shaded region is not convex.

• h:{(x, y) : 3x² + 4y² ≥ 5}

From the graph, it is evident that the solution set which is the grey shaded region is not convex.

• p:{(x, y) : y ≥ 2, y ≤ 4}

From the graph, the dark blue shaded region between the two bright lines is a convex set.

Hence, the solution is option D.

Given, X₁ and X₂ are optimal solutions of a Linear programming problem(LPP).

This means that, {X₁ ,X₂} C (a convex Set) as the optimal solution of a LPP is convex.

Now by using the definition of a Convex set,

A set of points C is called convex if, for all λ in the interval 0 ≤ λ ≤ 1, λy + (1 − λ)z is contained in C whenever y and z are contained in C.

By using this property of Convex set,

If {X₁ ,X₂} C (a convex set of optimal solutions), then

X = λX₁ + (1 − λ) X₂ where 0 ≤ λ ≤ 1, is also contained in C (the optimal solution set).

This proves that, also X C.

Hence the answer is option B.

Given,

Z = 4x + 2y

Subjected to constraints,

2x + 3y ≤ 18

x + y ≥ 10