Applications Of Derivatives

Let the side of the square be a

Rate of change of side=

Perimeter of the square = 4a

Rate of change of perimeter =

Let the radius of the circle be r

Circumference of the circle=

Rate of change of circumference

Let the radius of the circle be r

Area of the circle=

Rate of change of Area

Let the side of the square be a

Rate of change of side=

Area of the square = a²

Rate of change of Area =

Soap bubble will be in the shape of a sphere

Let the radius of the soap bubble be r

Surface area of the soap bubble=4

Rate of change of Surface area

Soap bubble will be in the shape of a sphere

Let the radius of the soap bubble be r

Volume of the soap bubble=

Rate of change of Volume

Let the radius of the balloon be r

Let the volume of the spherical balloon be V

Surface area of the bubble =4

Rate of change of Surface area

When we pump a balloon its volume changes.

Let the radius of the balloon be r

Let the volume of the water tank be V

Let the radius of the circle be r

Area of the circle=

Rate of change of Area

ABE and CDE are similar triangles.

So,

Let the volume of the cone be V

Let the volume of the cone be V

Let the original ladder be AC and the pulled ladder be DE

Let AB=y and BC=x

Applying Pythagoras Theorem in ABC

…(1)

Differentiating both sides of eqn (1) wrt to t

ATQ,

Let y =√x.

Let x =36 and Δx = 1.

As y = √x.

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.08

Also,

Δy = f(x+Δx)-f(x)

∴

⇒ 0.08 = √37 – 6

⇒ √37 = 6.08

Let .

Let x =27 and Δx = 2.

As

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.074

Also,

Δy = f(x+Δx)-f(x)

∴

⇒

⇒

Let y =√x.

Let x =25 and Δx = 2.

As y = √x.

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.2

Also,

Δy = f(x+Δx)-f(x)

∴

⇒ 0.2 = √27 – 5

⇒ √27 = 5.2

Let y =√x.

Let x =0.25 and Δx = -0.01.

As y = √x.

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = -0.01

Also,

Δy = f(x+Δx)-f(x)

∴

⇒ -0.01 = √0.24 – 0.5

⇒ √0.24 = 0.49

Let y =√x.

Let x =19 and Δx = 0.5.

As y = √x.

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.0357

Also,

Δy = f(x+Δx)-f(x)

∴

⇒ 0.0357 = √49.5 – 7

⇒ √49.5 = 7.0357.

Let .

Let x =16 and Δx = 1.

As

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = -0.03125

Also,

Δy = f(x+Δx)-f(x)

∴

⇒

⇒

Let

Let x =2 and Δx = 0.002.

As

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = -0.0005

Also,

Δy = f(x+Δx)-f(x)

∴

⇒

⇒

Let

Let x =10 and Δx = 0.02.

As

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.002

Also,

Δy = f(x+Δx)-f(x)

∴ 0.002 = log_e(10+0.02) – log_e(10)

⇒ 0.002 = log_e(10.02) – 2.3026

⇒ log_e(10.02) = 2.3046.

Let

∴

∴

Let x =4 and Δx = 0.04.

As

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.004343

Also,

Δy = f(x+Δx)-f(x)

∴ 0.004343 = log_e(4+0.04) – log_e(4)

⇒ 0.004343 = log_e(4.04) – 0.6021

⇒ log_e(4.04) = 0.606443.

Let y = cosx

Let x =60° and Δx = 1° .

As y =cos x

⇒

We, know

⇒

∴

⇒

⇒

∴ Δy = 0.01511

Also,

Δy = f(x+Δx)-f(x)

∴ 0.01511 = cos(60+1) – cos(60)

⇒ 0.01511 = cos61° – 0.5

⇒ cos61° = 0.48489

Given x is π /2

Value of π is 22/7

22/14 is π /2

Hence there will be no change.

Let the radius of the plate 10cm.

Radius increases by 2% by heating

∴ After increment =

Change in radius dr = 0.2

∴ New radius = 10+0.2 = 10.2cm

Area of circular plate = A=πr²

∴ Change in Area =

⇒

⇒

⇒

The formula for time period -

∴

Here 2,π,g have no dimensions. So we can eliminate them.

Now

By representing in percentage error

⇒

⇒

⇒

⇒

Hence the time period becomes 1 %.

Given: pv^1/4 = k

%decrease in the volume = 1/2%s

∴

pv^1/4 = k

taking log on both sides

log[pv^1/4] = log a

log P + 1.4logV = log a

Differentiating both the sides we get

⇒

⇒

Multiplying both sides by 100.

⇒

⇒

⇒

%error in P = 0.7%.

Decrease in radius = dr = 10-9.8

∴ dr = 0.2

Volume of the sphere is given by =

Change in volume =

∴

⇒

Surface area of the sphere is given by =

Change in volume =

∴ dA = 8π×10× 0.2

∴ dA = 16π.

Volume of the sphere is given by =

Change in volume =

Given: Δr = 0.1

⇒

⇒

Percentage error

⇒

= 0.3%

Let d be the diameter r be the radius and V be the volume of Sphere

Volume of the sphere is given by =

⇒

Let Δd be the error in d and the corresponding error in V be ΔV.

∴

∴

Hence Proved

Condition (1):

Since, f(x)=x² is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)=x² is continuous on [-1,1].

Condition (2):

Here, f’(x)=2x which exist in [-1,1].

So, f(x)=x² is differentiable on (-1,1).

Condition (3):

Here, f(-1)=(-1)²=1

And f(1)=1¹=1

i.e. f(-1)=f(1)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(-1,1) such that f’(c)=0

i.e. 2c=0

i.e. c=0

Value of c=0ϵ(-1,1)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x²-x-12 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x²-x-12 is continuous on [-3,4].

Condition (2):

Here, f’(x)=2x-1 which exist in [-3,4].

So, f(x)= x²-x-12 is differentiable on (-3,4).

Condition (3):

Here, f(-3)=(-3)²-3-12=0

And f(4)=4²-4-12=0

i.e. f(-3)=f(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(-3,4) such that f’(c)=0

i.e. 2c-1=0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=cosx is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x)=cosx is continuous on .

Condition (2):

Here, f’(x)=-sinx which exist in .

So, f(x)=cosx is differentiable on .

Condition (3):

Here,

And

i.e.

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one such that f’(c)=0

i.e. -sinc=0

i.e. c=0

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x²-5x+6 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x²-5x+6 is continuous on [2,3].

Condition (2):

Here, f’(x)=2x-5 which exist in [2,3].

So, f(x)= x²-5x+6 is differentiable on (2,3).

Condition (3):

Here, f(2)=2²-5×2+6=0

And f(3)= 3²-5×3+6=0

i.e. f(2)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(2,3) such that f’(c)=0

i.e. 2c-5=0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)= x²-5x+6 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x²-5x+6 is continuous on [-3,6].

Condition (2):

Here, f’(x)=2x-5 which exist in [-3,6].

So, f(x)= x²-5x+6 is differentiable on (-3,6).

Condition (3):

Here, f(-3)=(-3)²-5×(-3)+6=30

And f(6)= 6²-5×6+6=12

i.e. f(-3)≠f(6)

Conditions (3) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.

Condition (1):

Since, f(x)=x²-4x+3 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)=x²-4x+3 is continuous on [1,3].

Condition (2):

Here, f’(x)=2x-4 which exist in [1,3].

So, f(x)=x²-4x+3 is differentiable on (1,3).

Condition (3):

Here, f(1)=(1)²-4(1)+3=0

And f(3)= (3)²-4(3)+3=0

i.e. f(1)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(1,3) such that f’(c)=0

i.e. 2c-4=0

i.e. c=2

Value of c=2 ϵ(1,3)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x(x-4)² is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x(x-4)² is continuous on [0,4].

Condition (2):

Here, f’(x)= (x-4)²+2x(x-4) which exist in [0,4].

So, f(x)= x(x-4)² is differentiable on (0,4).

Condition (3):

Here, f(0)=0(0-4)²=0

And f(4)= 4(4-4)²=0

i.e. f(0)=f(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,4) such that f’(c)=0

i.e. (c-4)²+2c(c-4)=0

i.e. (c-4)(3c-4)=0

i.e. c=4 or c=3÷4

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x³- 7x²+16x-12 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x³- 7x²+16x-12 is continuous on [2,3].

Condition (2):

Here, f’(x)=3x²-14x+16 which exist in [2,3].

So, f(x)= x³- 7x²+16x-12 is differentiable on (2,3).

Condition (3):

Here, f(2)= 2³- 7(2)²+16(2)-12=0

And f(3)= 3³- 7(3)²+16(3)-12=0

i.e. f(2)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(2,3) such that f’(c)=0

i.e. 3c²-14c+16=0

i.e. (c-2)(3c-7)=0

i.e. c=2 or c=7÷3

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)= x³+3x²-24x-80 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x³+3x²-24x-80 is continuous on [-4,5].

Condition (2):

Here, f’(x)= 3x²+6x-24 which exist in [-4,5].

So, f(x)= x³+3x²-24x-80 is differentiable on (-4,5).

Condition (3):

Here, f(-4)= (-4)³+3(-4)²-24(-4)-80=0

And f(5)= (5)³+3(5)²-24(5)-80=0

i.e. f(-4)=f(5)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(-4,5) such that f’(c)=0

i.e. 3c²+6c-24=0

i.e. c=-4 or c=2

Value of c=2 ϵ(-4,5)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=(x-1)(x-2)(x-3) is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-1)(x-2)(x-3) is continuous on [1,3].

Condition (2):

Here, f’(x)= (x-2)(x-3)+ (x-1)(x-3)+ (x-1)(x-2) which exist in [1,3].

So, f(x)= (x-1)(x-2)(x-3) is differentiable on (1,3).

Condition (3):

Here, f(1)= (1-1)(1-2)(1-3) =0

And f(3)= (3-1)(3-2)(3-3) =0

i.e. f(1)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(1,3) such that f’(c)=0

i.e. (c-2)(c-3)+ (c-1)(c-3)+ (c-1)(c-2)=0

i.e. (c-3)(2c-3)+(c-1)(c-2)=0

i.e. (2c²-9c+9)+(c²-3c+2)=0

i.e. 3c²-12c+11=0

i.e.

i.e. c=2.58 or c=1.42

Value of c=1.42ϵ(1,3) and c=2.58ϵ(1,3)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=(x-1)(x-2)² is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-1)(x-2)² is continuous on [1,2].

Condition (2):

Here, f’(x)= (x-2)²+2(x-1)(x-2) which exist in [1,2].

So, f(x)= (x-1)(x-2)² is differentiable on (1,2).

Condition (3):

Here, f(1)= (1-1)(1-2)²=0

And f(2)= (2-1)(2-2)²=0

i.e. f(1)=f(2)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(1,2) such that f’(c)=0

i.e. (c-2)²+2(c-1)(c-2)=0

(3c-4)(c-2)=0

i.e. c=2 or c=4÷3

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=(x-2)⁴(x-3)³ is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-2)⁴(x-3)³ is continuous on [2,3].

Condition (2):

Here, f’(x)= 4(x-2)³(x-3)³+3(x-2)⁴(x-3)² which exist in [2,3].

So, f(x)= (x-2)⁴(x-3)³ is differentiable on (2,3).

Condition (3):

Here, f(2)= (2-2)⁴(2-3)³=0

And f(3)= (3-2)⁴(3-3)³=0

i.e. f(2)=f(3)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(2,3) such that f’(c)=0

i.e. 4(c-2)³(c-3)³+3(c-2)⁴(c-3)²=0

(c-2)³(c-3)²(7c-18)=0

i.e. c=2 or c=3 or c=18÷7

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ is continuous on [-1,1].

Condition (2):

Here, which exist in [-1,1].

So, is differentiable on (-1,1).

Condition (3):

Here,

And

i.e.

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(-1,1) such that f’(c)=0

i.e.

i.e. c=0

Value of c=0ϵ(-1,1)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=cos2x is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x)= cos2x is continuous on [0,π].

Condition (2):

Here, f’(x)= -2sin2x which exist in [0,π].

So, f(x)=cos2x is differentiable on (0,π).

Condition (3):

Here, f(0)=cos0=1

And f(π)=cos2π=1

i.e. f(0)=f(π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,π) such that f’(c)=0

i.e. -2sin2c =0

i.e. 2c=π

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=sin3x is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x)= sin3x is continuous on [0,π].

Condition (2):

Here, f’(x)= 3cos3x which exist in [0,π].

So, f(x)= sin3x is differentiable on (0,π).

Condition (3):

Here, f(0)=sin0=0

And f(π)=sin3π=0

i.e. f(0)=f(π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,π) such that f’(c)=0

i.e. 3cos3c =0

i.e.

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=sinx+cosx is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x)= sinx+cosx is continuous on .

Condition (2):

Here, f’(x)= cosx-sinx which exist in .

So, f(x)= sinx+cosx is differentiable on

Condition (3):

Here, f(0)=sin0+cos0=1

And

i.e.

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one such that f’(c)=0

i.e. cosc-sinc =0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=e^-x sinx is a combination of exponential and trigonometric function which is continuous.

⇒ f(x)= e^-x sinx is continuous on [0,π].

Condition (2):

Here, f’(x)= e^-x (cosx – sinx) which exist in [0,π].

So, f(x)= e^-x sinx is differentiable on (0,π)

Condition (3):

Here, f(0)= e^-0 sin0=0

And f(π)= e^-πsinπ =0

i.e. f(0)=f(π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,π) such that f’(c)=0

i.e. e^-c (cos c – sin c) =0

i.e. cos c-sin c = 0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=e^-x (sinx-cosx) is a combination of exponential and trigonometric function which is continuous.

⇒ f(x)= e^-x (sinx-cosx) is continuous on .

Condition (2):

Here, f’(x)= e^-x (sinx + cosx) - e^-x (sinx – cosx)

= e^-x cosx which exist in .

So, f(x)= e^-x (sinx-cosx) is differentiable on

Condition (3):

Here,

And

i.e.

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one such that f’(c)=0

i.e. e^-c cos c =0

i.e. cos c = 0

i.e.

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x) = sinx-sin2x is a trigonometric function and we know every trigonometric function is continuous.

⇒ f(x) = sinx-sin2x is continuous on [0,2π].

Condition (2):

Here, f’(x)= cosx-2cos2x which exist in [0,2π].

So, f(x)= sinx-sin2x is differentiable on (0,2π)

Condition (3):

Here, f(0)= sin0-sin0 = 0

And f(2π)=sin(2π)-sin(4π) =0

i.e. f(0)=f(2π)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,2π) such that f’(c)=0

i.e. cosx-2cos2x =0

i.e. cosx-4cos²x+2=0

i.e. 4cos²x-cosx-2=0

i.e.

i.e. c=32° 32’ or c=126°23’

Value of c=32°32’ϵ(0,2π)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x(x+2)e^x is a combination of exponential and polynomial function which is continuous for all xϵR.

⇒ f(x)= x(x+2)e^x is continuous on [-2,0].

Condition (2):

Here, f’(x)=(x²+4x+2)e^x which exist in [-2,0].

So, f(x)=x(x+2)e^x is differentiable on (-2,0).

Condition (3):

Here, f(-2)= (-2)(-2+2)e^-2 =0

And f(0)= 0(0+2)e⁰=0

i.e. f(-2)=f(0)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(-2,0) such that f’(c)=0

i.e. (c²+4c+2)e^c =0

i.e. (c+√2)²=0

i.e. c=-√2

Value of c=-√2 ϵ(-2,0)

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=x(x-5)² is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x(x-5)² is continuous on [0,5].

Condition (2):

Here, f’(x)= (x-5)²+ 2x(x-5) which exist in [0,5].

So, f(x)= x(x-5)² is differentiable on (0,5).

Condition (3):

Here, f(0)= 0(0-5)²=0

And f(5)= 5(5-5)²=0

i.e. f(0)=f(5)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,5) such that f’(c)=0

i.e. (c-5)²+ 2c(c-5)=0

i.e.(c-5)(3c-5)=0

i.e. or c=5

Value of

Thus, Rolle’s theorem is satisfied.

Condition (1):

Since, f(x)=(x-1)(2x-3) is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= (x-1)(2x-3) is continuous on [1,3].

Condition (2):

Here, f’(x)= (2x-3)+ 2(x-1) which exist in [1,3].

So, f(x)= (x-1)(2x-3) is differentiable on (1,3).

Condition (3):

Here, f(1)= (1-1)(2(1)-3)=0

And f(5)= (3-1)(2(3)-3)=6

i.e. f(1)≠f(3)

Condition (3) of Rolle’s theorem is not satisfied.

So, Rolle’s theorem is not applicable.

Condition (1):

Since, f(x)=x^1/2 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= x^1/2 is continuous on [-1,1].

Condition (2):

Here, which does not exist at x=0 in [-1,1].

f(x)=x^1/2 is not differentiable on (-1,1).

Condition (2) of Rolle’s theorem is not satisfied.

So,Rolle’s theorem is not applicable.

Condition (1):

Since, f(x)=2+(x-1)^2/3 is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ f(x)= 2+(x-1)^2/3 is continuous on [0,2].

Condition (2):

Here, which does not exist at x=1 in [0,2].

f(x)= 2+(x-1)^2/3 is not differentiable on (0,2).

Condition (2) of Rolle’s theorem is not satisfied.

So,Rolle’s theorem is not applicable.

Condition (1):

Since, which is discontinuous at x=0

is not continuous on [-1,1].

Condition (1) of Rolle’s theorem is not satisfied.

So,Rolle’s theorem is not applicable.

Condition (1):

Since, f(x)=[x] which is discontinuous at x=0

⇒ f(x)=[x] is not continuous on [-1,1].

Condition (1) of Rolle’s theorem is not satisfied.

So,Rolle’s theorem is not applicable.

Condition (1):

Since, y=x(x-4) is a polynomial and we know every polynomial function is continuous for all xϵR.

⇒ y= x(x-4) is continuous on [0,4].

Condition (2):

Here, y’= (x-4)+x which exist in [0,4].

So, y= x(x-4) is differentiable on (0,4).

Condition (3):

Here, y(0)=0(0-4)=0

And y(4)= 4(4-4)=0

i.e. y(0)=y(4)

Conditions of Rolle’s theorem are satisfied.

Hence, there exist at least one cϵ(0,4) such that y’(c)=0

i.e. (c-4)+c=0

i.e. 2c-4=0

i.e. c=2

Value of c=2ϵ(0,4)

So,y(c)=y(2)=2(2-4)=-4

By geometric interpretation, (2,-4) is a point on a curve y=x(x-4),where tangent is parallel to x-axis.

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

=12

⇒ f' (c)=2c+2

⇒ 2c+2=12

⇒ c=5

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [0,4].

=5

⇒ f'(c)=2c+1

⇒ 2c+1=5

⇒ c=2

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [1,3].

=5

⇒ f'(c)=4c-3

⇒ 4c-3=5

⇒ c=2

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [-1,4].

=10

f' (c)=3c²+2c-6

⇒ 3 c²+2c-6=10

⇒ 3 c²+2c-16=0

⇒ 3 c²-6c+8c-16=0

⇒ 3c(c-2)+8(c-2)=0

⇒ (3c+8)(c-2)=0

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

=0

⇒ f' (c)=3c²-36c+104

=3c²-36c+10

=0

Given:

Since f(c) is continuous as well as differentiable in the interval [0,1].

⇒ f' (c)=e^c

⇒ e^c =e-1

⇒ c=log_e(e-1)

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [0,1].

=1

Given:

Since log x is a continuous as well as differentiable function in the interval [1,e].

c=e-1

Given:

Since is a continuous as well as differentiable function in the interval [0,1].

Given:

Since sin x is a continuous as well as differentiable function in the interval .

=0

f' (c)=cos x

cos x=0

Given:

Since (sin x + cos x) is a continuous as well as differentiable function in the interval .

=0

f' (c)=cos x-sin x

⇒ cos x-sin x=0

Given:

Since f(x) is continuous in the interval [-1,1].

But is non differentiable at x=0 due to sharp corner.

So LMVT is not applicable to ƒ(x)=|x|

Given:

Since the graph is discontinuous at x=0 as shown in the graph.

So LMVT is not applicable to the above function.

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [0,].

f' (c)=3x² -6x+2

3 x²-6x+2=3/4

12 x²-24x+8=3

12 x²-24x+5=0

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [1,5].

)

⇒ 16c²=600-24c²

⇒ 40c²=600

⇒ c²=15

Given:

Since the f(x) is a polynomial function,

It is continuous as well as differentiable in the interval [4,6].

⇒ c=4.964

Given:

y=x²

Since y is a polynomial function.

It is continuous and differentiable in [1,2]

So, there exists a c such that:

⇒ f' (c)=2c

⇒ 2c=3

So, the point is

Given:

Since y is a polynomial function.

It is continuous and differentiable in [1,3]

So, there exists a c such that:

⇒ f' (c)=3c²

⇒ 3c²=13

So the point is

Given:

y=x³-3x

Since y is a polynomial function.

It is continuous and differentiable in [1,2]

So, there exists a c such that:

=4

⇒ f' (c)=3c²-3

⇒ 3 c²-3=4

⇒ 3c²=7

So, the points are

Given:

f(x)=x(1-log x)

Since the function is continuous as well as differentiable

So, there exists c such that

(b-a) log c=b(1-log b )-a(1-log a)

Hence proved.

min. value = 4

Since the square of any no. Is positive, the given function has no maximum value.

The minimum value exists when the quantity (5x-1)²=0

Therefore, minimum value=4

max. value = 9

Since the quantity (x-3)² has a –ve sign, the max. Value it can have is 9.

Also hence it has no minimum value.

max. value = 6

Since |x+4| is non-negative for all x belonging to R.

Therefore the least value it can have is 0 .

Hence value of function is 6.

It has no minimumvalue as it can have infinitely many.

max. value = 4, min. value = 6

f(x)=sin2x+5

We know that,

-1≤sinӨ≤1

-1≤sin2x≤1

Adding 5 on both sides,

-1+5≤sin2x+5≤1+5

4≤sin2x+5≤6

Hence

max value of f(x)=2x+5 will be 6

Min value of f(x) =2x+5 will be 4

max. value = 4, min. value = 2

We know that

-1≤sinӨ≤1

-1≤sin4x≤1

Adding 3 on both sides,

We get

-1+3≤sin4x+3≤1+3

2≤|sin4x+3|≤4

Hence min.Value is 2 and max value is 4

local max. value is 0 at x = 3

F^’(x)=4(x-3)³=0

⇒X=3

؞ local max. Vaue is 0.

local min. value is 0 at x = 0

F^’(x)=2x=0

x=0

؞ local min.value is 0

local max. value is −3 at x = 1 and local min. value is −128 at x = 6

F^’(x)=6x²-42x+36=0

⇒ 6(x-1)(x-6)=0

⇒x=1,6

F^’’(x)=12x-42

F^’’(1)<0 ,1 is the pont of local max.

F^’’(6)>0, 6 is the point of localmin.

F(1)=2-21+36-20=-3

F(6)=-128

local max. value is 19 at x = 1 and local min. value is 15 at x = 3

F^’(x)=3x²-12x+9=0

⇒3(x-3)(x-1)=0

⇒ x=3,1

F^’’(x)=6x-12

F^’’(3)=18-12=6>0 , 3 is the of local min.

F^’’(1)<0, 1 is the point of local max.

F(3)=15

F(1)=19

local max. value is 68 at x = 1 and local min. values are −1647 at x = −6 and −316 at x = 5

F’(x)=4x³-124x+120=0

⇒ 4(x³-31x+30)=0

For x=1, the given eq is 0

؞x-1 is a factor,

4(x-1)(x+6)(x-5)=0

⇒X=1,-6,5

F^’’(1)<0, 1is the point of max.

F^’’(-6) and f^’’(5)>0, -6 and 5 are point of min.

F(1)=68

F(-6)=-1647

F(5)=-316

local max. value is 251 at x = 8 and local min. value is −5 at x = 0

‘^}(x)=-3x²+24x=0

⇒ -3x(x-8)=0

⇒ x=0,8

F^’’(x)=-6x+24

F^’’(0)>0, 0 is the point of local min.

F^’’(8)<0, 8 is the point of local max.

F(8)=251 and f(0)=-5

local max. value is 0 at x = −2 and local min. value is −4 at x = 0

f^’(x)=(x-1)2(x+2)+(x+2)²=0

x=0,-2

f^’’(0)>0, 0 is the point of local min.

f^’’(-2)<0, -2 is the point of local max.

f(0)=-4

f(-2)=0

local max. value is 0 at each of the points x = 1 and x = −1 and local min. value is at

F^’(x)=-(x-1)³2(x+1)-3(x-1)²(x+1)²=0

Since, f^||(1) and f^||(-1) <0 , 1 and -1 are the points of local max.

F^||(-)>0, - is the point of local min.

F(1)=f(-1)=0

Also,

local min. value is 2 at x = 2

F^’(x)=

⇒ x²-4=0

⇒ x=±2

But since x>0, x=2

؞point of local mini. is 2

F(2)=

max. value is 139 at x = −2 and min. value is 89 at x = 3

F^’(x)=6x²-24=0

6(x²-4)=0

6(x²-2²)=0

6(x-2)(x+2)=0

X=2,-2

Now, we shall evaluate the value of f at these points and the end points

F(2)=2(2)³-24(2)+107=75

F(-2)=2(-2)³-24(-2)+107=139

F(-3)=2(-3)³-24(-3)+107=125

F(3)=2(3)³-24(3)+107=89

max. value is 257 at x = 4 and min. value is −63 at x = 2

F^|(x)=12x³-24x²+24x-48=0

12(x³-2x²+2x-4)=0

Since for x=2, x³-2x²+2x-4=0, x-2 is a factor

On dividing x³-2x²+2x-4 by x-2, we get,

12(x-2)(x²+2)=0

X=2,4

Now, we shall evaluate the value of f at these points and the end points

F(1)=3(1)⁴-8(1)³+12(1)²-48(1)+1=-40

F(2)= 3(2)⁴-8(2)³+12(2)²-48(2)+1=-63

F(4)= 3(4)⁴-8(4)³+12(4)²-48(4)+1=257

max. value is at and min. value is at

F^|(x)=cos x-sin x=0

_ð 2 cos x=sin x

The given function is

Now, taking logarithm from both sides, we get..

Differentiating both sides w.r.t x….

(1-ln(x))=0

ln(x)=1

x=e

hence the max. point is x=e

max value is .

F(x)=x+

Taking first derivative and equating it to zero to find extreme points.

F^’(x)=1-

X²=1

x=1,x=-1

now to determine which of these is min. And max. We use second derivative.

f^||(x)=

f^||(1)=2 and f^||(-1)=-2

since f^||(1) is +ve it is minimum point while f^||(-1) is –ve it is maximum point

max value-> f(-1)=-1+=-2

min vaue-> f(1)=1+=2

hence maximum value is less than minimum value

49

=0

⇒x=−23

Step 2

=−36 is negative

Step 3

maximum profit

=49

(1, 3)

Let P(x,y) be the position of the jet and the soldier is placed at A(3,2)

AP=

As y=x²+2 or y-2=x²

؞ AP²=(x-3)² + x⁴=z (say)

؞ 2x-6+4x³=0

Put x=1

2-6+4=0

؞ x-1 is a factor

And

or x=1

(at x=1)>0

؞ z is minimum when x=1,y=1+2=3

Point is (1,3)

max. value is at and min. value is at

f^’(x)=-1+2cos x=0

By finding the general solution, we get and

Now, by finding the second derivative, we get that and

Therefore, max. value is at and min. value is at

Given,

• The two numbers are positive.

• the product of two numbers is 49.

• the sum of the two numbers is minimum.

Let us consider,

• x and y are the two numbers, such that x > 0 and y > 0

• Product of the numbers : x × y = 49

• Sum of the numbers : S = x + y

Now as,

x × y = 49

------ (1)

Consider,

S = x + y

By substituting (1), we have

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with x

----- (3)

[Since and ]

Now equating the first derivative to zero will give the critical point c.

So,

= x² = 49

As x > 0, then x = 7

Now, for checking if the value of S is maximum or minimum at x=7, we will perform the second differentiation and check the value ofat the critical value x = 7.

Performing the second differentiation on the equation (3) with respect to x.

[Since and ]

Now when x = 7,

As second differential is positive, hence the critical point x = 7 will be the minimum point of the function S.

Therefore, the function S = sum of the two numbers is minimum at x = 7.

From Equation (1), if x= 7

Therefore, x = 7 and y = 7 are the two positive numbers whose product is 49 and the sum is minimum.

Given,

• The two numbers are positive.

• the sum of two numbers is 16.

• the sum of the squares of two numbers is minimum.

Let us consider,

• x and y are the two numbers, such that x > 0 and y > 0

• Sum of the numbers : x + y = 16

• Sum of squares of the numbers : S = x² + y²

Now as,

x + y = 16

y = (16-x) ------ (1)

Consider,

S = x² + y²

By substituting (1), we have

S = x² + (16-y)² ------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with x

----- (3)

[Since ]

Now equating the first derivative to zero will give the critical point c.

So,

⇒ 2x – 2(16 -x) = 0

⇒ 2x – 32 + 2x = 0

= 4x = 32

⇒ x = 8

As x > 0, x = 8

Now, for checking if the value of S is maximum or minimum at x=8, we will perform the second differentiation and check the value ofat the critical value x = 8.

Performing the second differentiation on the equation (3) with respect to x.

[Since and ]

Now when x = 8,

As second differential is positive, hence the critical point x = 8 will be the minimum point of the function S.

Therefore, the function S = sum of the squares of the two numbers is minimum at x = 8.

From Equation (1), if x= 8

y = 16 – 8 = 8

Therefore, x = 8 and y = 8 are the two positive numbers whose su is 16 and the sum of the squares is minimum.

Given,

• the number 15 is divided into two numbers.

• the product of the square of one number and cube of another number is maximum.

Let us consider,

• x and y are the two numbers

• Sum of the numbers : x + y = 15

• Product of square of the one number and cube of anther number : P = x³ y²

Now as,

x + y = 15

y = (15-x) ------ (1)

Consider,

P = x³y²

By substituting (1), we have

P = x³ × (15-x)² ------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with x

[Since and if u and v are two functions of x, then ]

= 3×[15² – 2× (15)×(x) + x²] x² + x³(2x-30)

= x²[3× (225 – 30x + x²)+ x (2x - 30)]

= x²[ 675– 90x + 3x²+ 2x² – 60x]

= x²[5x² – 120x + 675]

= 5x² [x² – 24x + 135] ----- (3)

Now equating the first derivative to zero will give the critical point c.

So,

Hence 5x² = 0 (or) x² – 24x + 135 = 0

x = 0 (or)

x = 0 (or)

x = 0 (or)

x = 0 (or)

x = 0 (or) (or)

x = 0 (or) (or)

x = 0 (or) x = 15 (or) x = 9

Now considering the critical values of x = 0,9,15

Now, for checking if the value of P is maximum or minimum at x=0,9,15, we will perform the second differentiation and check the value ofat the critical value x = 0,9,15.

Performing the second differentiation on the equation (3) with respect to x.

= (x² – 24x + 135) (5 × 2x) + 5x² (2x – 24 + 0)

[Since and and if u and v are two functions of x, then ]

= (x² – 24x + 135) (10x) + 5x² (2x – 24)

= 10x³ – 240x² + 1350x + 10x³ – 120x²

= 20x³ – 360x² + 1350x

= 5x (4x² – 72x + 270)

Now when x = 0,

= 0

So, we reject x = 0

Now when x = 15,

= 65 [(4 × 225) –1080+ 270]

= 65 [900– 1080+ 270]

= 65 [1170– 1080]

= 65× (90) > 0

Hence , so at x = 15, the function P is minimum

Now when x = 9,

= 45 [(4 × 81) – 648 + 270]

= 45 [324 – 648 + 270]

= 45 [594 – 648]

= 45 × (-54)

= -2430 < 0

As second differential is negative, hence at the critical point x = 9 will be the maximum point of the function P.

Therefore, the function P is maximum at x = 9.

From Equation (1), if x= 9

y = 15 – 9 = 6

Therefore, x = 9 and y = 6 are the two positive numbers whose sum is 15 and the product of the square of one number and cube of another number is maximum.

Given,

• the number 8 is divided into two numbers.

• the product of the square of one number and cube of another number is minimum.

Let us consider,

• x and y are the two numbers

• Sum of the numbers : x + y = 8

• Product of square of the one number and cube of anther number : S = x³ + y²

Now as,

x + y = 8

y = (8-x) ------ (1)

Consider,

S = x³ + y²

By substituting (1), we have

S = x³ + (8-x)² ------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with x

[Since ]

= 3x² + 2x – 16 ------ (3)

Now equating the first derivative to zero will give the critical point c.

So,

Hence 3x² + 2x - 16= 0

(or)

(or)

x = 2 (or) x = -2.67

Now considering the critical values of x = 2,-2.67

Now, for checking if the value of P is maximum or minimum at x=2,-2.67, we will perform the second differentiation and check the value ofat the critical value x = 2,-2.67.

Performing the second differentiation on the equation (3) with respect to x.

= 3 (2x) + 2 (1) - 0

[Since and ]

= 6x + 2

Now when x = -2.67,

= - 16.02 + 2 = -14.02

At x = -2.67 hence, the function S will be maximum at this point.

Now consider x = 2,

= 12 + 2 = 14

Hence , so at x = 2, the function S is minimum

As second differential is positive, hence at the critical point x = 2 will be the maximum point of the function S.

Therefore, the function S is maximum at x = 2.

From Equation (1), if x= 2

y = 8 – 2 = 6

Therefore, x = 2 and y = 6 are the two positive numbers whose sum is 8 and the sum of the square of one number and cube of another number is maximum.

Given,

• the number ‘a’ is divided into two numbers.

• the product of the pth power of one number and qth power of another number is maximum.

Let us consider,

• x and y are the two numbers

• Sum of the numbers : x + y = a

• Product of square of the one number and cube of anther number : P = x^p y^q

Now as,

x + y = a

y = (a-x) ------ (1)

Consider,

P = x^py^q

By substituting (1), we have

P = x^p × (a-x)^q ------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with x

[Since and if u and v are two functions of x, then ]

= x^p-1(a-x)^q-1[ap-xp-xq]

= x^p-1(a-x)^q-1[ap – x (p+q)] ----- (3)

Now equating the first derivative to zero will give the critical point c.

So,

Hence x^p-1 = 0 (or) (a-x)^q-1 (or) ap– x(p+q)= 0

x = 0 (or) x = a (or)

Now considering the critical values of x = 0,a and

Now, using the First Derivative test,

For f, a continuous function which has a critical point c, then, function has the local maximum at c, if f’(x) changes the sign from positive to negative as x increases through c, i.e. f’(x)>0 at every point close to the left of c and f’(x)<0 at every point close to the right of c.

Now when x = 0,

So, we reject x = 0

Now when x = a,

Hence we reject x = a

Now when ,

> 0 ---- (4)

Now when ,

< 0 ---- (5)

By using first derivative test, from (4) and (5), we can conclude that, the function P has local maximum at

From Equation (1), if

Therefore, and are the two positive numbers whose sum together to give the number ‘a’ and whose product of the pth power of one number and qth power of the other number is maximum.

Given:

Rate of working of an engine R, v is the speed of the engine:

, where 0<v<30

For finding the maximum/ minimum of given function, we can find it by differentiating it with v and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Now, differentiating the function R with respect to v.

----- (1)

[Since and ]

Equating equation (1) to zero to find the critical value.

v² = 400

v = 20 (or) v = -20

As given in the question 0<v<30, v = 20

Now, for checking if the value of R is maximum or minimum at v=20, we will perform the second differentiation and check the value ofat the critical value v = 20.

Differentiating Equation (1) with respect to v again:

[Since and ]

----- (2)

Now find the value of

So, at critical point v = 20. The function R is at its minimum.

Hence, the function R is at its minimum at v = 20.

Given,

• Area of the rectangle is 93 cm².

• The perimeter of the rectangle is also fixed.

Let us consider,

• x and y be the lengths of the base and height of the rectangle.

• Area of the rectangle = A = x × y = 96 cm²

• Perimeter of the rectangle = P = 2 (x + y)

As,

x × y = 96

------ (1)

Consider the perimeter function,

P = 2 (x + y)

Now substituting (1) in P,

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since and ]

----- (3)

To find the critical point, we need to equate equation (3) to zero.

As the length and breadth of a rectangle cannot be negative, hence

Now to check if this critical point will determine the least perimeter, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with x:

[Since and ]

------ (4)

Now, consider the value of

As , so the function P is minimum at .

Now substituting in equation (1):

[By rationalizing he numerator and denominator with ]

Hence, area of the rectangle with sides of a rectangle with is 96cm² and has the least perimeter.

Now the perimeter of the rectangle is

The least perimeter is .

Given,

• Rectangle with given perimeter.

Let us consider,

• ‘p’ as the fixed perimeter of the rectangle.

• ‘x’ and ‘y’ be the sides of the given rectangle.

• Area of the rectangle, A = x × y.

Now as consider the perimeter of the rectangle,

p = 2(x +y)

p = 2x + 2y

----- (1)

Consider the area of the rectangle,

A = x × y

Substituting (1) in the area of the rectangle,

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

----- (3)

To find the critical point, we need to equate equation (3) to zero.

Now to check if this critical point will determine the largest rectangle, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

[Since and ]

------ (4)

Now, consider the value of

As , so the function P is maximum at .

Now substituting in equation (1):

As the sides of the taken rectangle are equal, we can clearly say that a largest rectangle which has a given perimeter is a square.

Given,

• Rectangle with given perimeter.

Let us consider,

• ‘p’ as the fixed perimeter of the rectangle.

• ‘x’ and ‘y’ be the sides of the given rectangle.

• Diagonal of the rectangle, . (using the hypotenuse formula)

Now as consider the perimeter of the rectangle,

p = 2(x +y)

p = 2x + 2y

----- (1)

Consider the diagonal of the rectangle,

Substituting (1) in the diagonal of the rectangle,

[squaring both sides]

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

= 2x - p + 2x

------ (3)

To find the critical point, we need to equate equation (3) to zero.

4x –p = 0

4x = p

Now to check if this critical point will determine the minimum diagonal, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with x:

= 4 + 0

[Since and ]

------ (4)

Now, consider the value of

As , so the function Z is minimum at .

Now substituting in equation (1):

As the sides of the taken rectangle are equal, we can clearly say that a rectangle with minimum diagonal which has a given perimeter is a square.

Given,

• Rectangle is of maximum perimeter.

• The rectangle is inscribed inside a circle.

• The radius of the circle is ‘a’.

Let us consider,

• ‘x’ and ‘y’ be the length and breadth of the given rectangle.

• Diagonal AC² = AB² + BC² is given by 4a² = x²+y² (as AC = 2a)

• Perimeter of the rectangle, P = 2(x+y)

Consider the diagonal,

4a² = x² + y²

y² = 4a² – x²

---- (1)

Now, perimeter of the rectangle, P

P = 2x + 2y

Substituting (1) in the perimeter of the rectangle.

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

[squaring on both sides]

4a² – x² = x²

2x² = 4a²

x² = 2a²

[as x cannot be negative]

Now to check if this critical point will determine the maximum diagonal, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

[Since and and if u and v are two functions of x, then ]

----- (4)

Now, consider the value of

As , so the function P is maximum at .

Now substituting in equation (1):

As the sides of the taken rectangle are equal, we can clearly say that a rectangle with maximum perimeter which is inscribed inside a circle of radius ‘a’ is a square.

Given,

• Sum of perimeter of square and circle.

Let us consider,

• ‘x’ be the side of the square.

• ‘r’ be the radius of the circle.

• Let ‘p’ be the sum of perimeters of square and circle.

p = 4x + 2πr

Consider the sum of the perimeters of square and circle.

p = 4x + 2πr

4x = p – 2πr

---- (1)

Sum of the area of the circle and square is

A = x² + πr²

Substituting (1) in the sum of the areas,

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with r and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to r:

[Since and ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

Now to check if this critical point will determine the least of the sum of the areas of square and circle, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with r:

[Since and ]

----- (4)

Now, consider the value of

As , so the function A is minimum at .

Now substituting in equation (1):

As the side of the square,

[as ]

Therefore, side of the square, x = 2r = diameter of the circle.

Given,

• A right angle triangle is inscribed inside the circle.

• The radius of the circle is given.

Let us consider,

• ‘r’ is the radius of the circle.

• ‘x’ and ‘y’ be the base and height of the right angle triangle.

• The hypotenuse of the ΔABC = AB² = AC² + BC²

AB = 2r, AC = y and BC = x

Hence,

4r² = x² + y²

y² = 4r² – x²

--- (1)

Now, Area of the ΔABC is

Now substituting (1) in the area of the triangle,

[Squaring both sides]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since and if u and v are two functions of x, then ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

[as the base of the triangle cannot be negative.]

Now to check if this critical point will determine the maximum area of the triangle, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now, consider the value of

As , so the function A is maximum at .

Now substituting in equation (1):

As , the base and height of the triangle are equal, which means that two sides of a right angled triangle are equal,

Hence the given triangle, which is inscribed in a circle, is an isosceles triangle with sides AC and BC equal.

Given,

• A right angle triangle.

• Hypotenuse of the given triangle is given.

Let us consider,

• ‘h’ is the hypotenuse of the given triangle.

• ‘x’ and ‘y’ be the base and height of the right angle triangle.

• The hypotenuse of the ΔABC = AC² = AB² + BC²

AC = h, AB = x and BC = y

Hence,

h² = x² + y²

y² = h² – x²

--- (1)

Now, perimeter of the ΔABC is

P = h + x + y

Now substituting (1) in the area of the triangle,

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

[squaring on both sides]

[as the base of the triangle cannot be negative.]

Now to check if this critical point will determine the maximum perimeter of the triangle, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

[Since if u and v are two functions of x, then ]

Now, consider the value of

As , so the function A is maximum at .

Now substituting in equation (1):

As , the base and height of the triangle are equal, which means that two sides of a right angled triangle are equal,

Hence the given triangle is an isosceles triangle with sides AB and BC equal.

Given,

• Perimeter of a triangle is 8 cm.

• One of the sides of the triangle is 3 cm.

• The area of the triangle is maximum.

Let us consider,

• ‘x’ and ‘y’ be the other two sides of the triangle.

Now, perimeter of the ΔABC is

8 = 3 + x + y

y = 8-3-x = 5-x

y = 5-x --- (1)

Consider the Heron’s area of the triangle,

Where

As perimeter = a + b+ c = 8

Now Area of the triangle is given by

Now substituting (1) in the area of the triangle,

[squaring on both sides]

Z = A² = 4(5x –x²-4) ----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

20 – 8x = 0

8x = 20

Now to check if this critical point will determine the maximum area of the triangle, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

As , so the function A is maximum at .

Now substituting in equation (1):

y = 5 – 2.5

y = 2.5

As x = y = 2.5, two sides of the triangle are equal,

Hence the given triangle is an isosceles triangle with two sides equal to 2.5 cm and the third side equal to 3cm.

Given,

• Window is in the form of a rectangle which has a semicircle mounted on it.

• Total Perimeter of the window is 10 metres.

• The total area of the window is maximum.

Let us consider,

• The breadth and height of the rectangle be ‘x’ and ‘y’.

• The radius of the semicircle will be half of the base of the rectangle.

Given Perimeter of the window is 10 meters:

[as the perimeter of the window will be equal to one side (x) less to the perimeter of rectangle and the perimeter of the semicircle.]

From here,

----- (1)

Now consider the area of the window,

Area of the window = area of the semicircle + area of the rectangle

Substituting (1) in the area equation:

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

------ (3)

To find the critical point, we need to equate equation (3) to zero.

x (4 +π) = 20

Now to check if this critical point will determine the maximum area of the window, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

[Since ]

------ (4)

As , so the function A is maximum at
.

Now substituting in equation (1):

Hence the given window with maximum area has breadth, and height, .

Given,

• Side of the square piece is 12 cms.

• the volume of the formed box is maximum.

Let us consider,

• ‘x’ be the length and breadth of the piece cut from each vertex of the piece.

• Side of the box now will be (12-2x)

• The height of the new formed box will also be ‘x’.

Let the volume of the newly formed box is :

V = (12-2x)² × (x)

V = (144 + 4x² – 48x) x

V = 4x³ -48x² +144x ------ (1)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (1) with respect to x:

-------- (2)

[Since ]

To find the critical point, we need to equate equation (2) to zero.

x² – 8x +12 = 0

x = 6 or x =2

x= 2

[as x = 6 is not a possibility, because 12-2x = 12-12= 0]

Now to check if this critical point will determine the maximum area of the box, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now let us find the value of

As , so the function A is maximum at x = 2

Now substituting x = 2 in 12 – 2x, the side of the considered box:

Side = 12-2x = 12 - 2(2) = 12-4= 8cms

Therefore side of wanted box is 8cms and height of the box is 2cms.

Now, the volume of the box is

V = (8)² × 2 = 64 × 2 = 128cm³

Hence maximum volume of the box formed by cutting the given 12cms sheet is 128cm³ with 8cms side and 2cms height.

Given,

• The open box has a square base

• The area of the box is c² square units.

• The volume of the box is maximum.

Let us consider,

• The side of the square base of the box be ‘a’ units. (pink coloured in the figure)

• The breadth of the 4 sides of the box will also be ‘a’units (skin coloured part).

• The depth of the box or the length of the sides be ‘h’ units (skin coloured part).

Now, the area of the box =

(area of the base) + 4 (area of each side of the box)

So as area of the box is given c²,

c² = a² + 4ah

---- (1)

Let the volume of the newly formed box is :

V = (a)² × (h)

[substituting (1) in the volume formula]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with a and then equating it to zero. This is because if the function f(a) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to a:

------- (3)

[Since ]

To find the critical point, we need to equate equation (3) to zero.

c² – 3a² = 0

[as ‘a’ cannot be negative]

Now to check if this critical point will determine the maximum Volume of the box, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with x:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at

Now substituting a in equation (1)

Therefore side of wanted box has a base side, is and height of the box, .

Now, the volume of the box is

Given,

• The can is cylindrical with a circular base

• The volume of the cylinder is 1 litre = 1000 cm².

• The surface area of the box is minimum as we need to find the minimum dimensions.

Let us consider,

• The radius base and top of the cylinder be ‘r’ units. (skin coloured in the figure)

• The height of the cylinder be ‘h’units.

• As the Volume of cylinder is given, V = 1000cm³

The Volume of the cylinder= πr²h

1000 = πr²h

---- (1)

The Surface area cylinder is = area of the circular base + area of the circular top + area of the cylinder

S = πr² + πr² + 2πrh

S = 2 πr² + 2πrh

[substituting (1) in the volume formula]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with r and then equating it to zero. This is because if the function f(r) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to r:

[Since and ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

Now to check if this critical point will determine the minimum surface area of the box, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with r:

----- (4)

[Since and ]

Now let us find the value of

As , so the function S is minimum at

Now substituting r in equation (1)

Therefore the radius of base of the cylinder, and height of the cylinder, where the surface area of the cylinder is minimum.

Given,

• The volume of the cone.

• The cone is right circular cone.

• The cone has least curved surface.

Let us consider,

• The radius of the circular base be ‘r’ cms.

• The height of the cone be ‘h’ cms.

• The slope of the cone be ‘l’ cms.

Given the Volume of the cone = πr²l

----- (1)

The Surface area cylinder is = πrl

S = πrl

[substituting (1) in the Surface area formula]

[squaring on both sides]

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with r and then equating it to zero. This is because if the function Z has a maximum/minimum at a point c then Z’(c) = 0.

Differentiating the equation (2) with respect to r:

[Since and ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

---- (4)

Now to check if this critical point will determine the minimum surface area of the cone, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with r:

[Since and ]

Now let us find the value of

As , so the function Z = S² is minimum

Now consider, the equation (4),

Now substitute the volume of the cone formula in the above equation.

π²r⁴h² = 2 π²r⁶

2r² = h²

Hence, the relation between h and r of the cone is proved when S is the minimum.

Given,

• The closed is cylindrical can with a circular base and top.

• The volume of the cylinder is 1 litre = 100 cm³.

• The surface area of the box is minimum.

Let us consider,

• The radius base and top of the cylinder be ‘r’ units. (skin coloured in the figure)

• The height of the cylinder be ‘h’units.

• As the Volume of cylinder is given, V = 100cm³

The Volume of the cylinder= πr²h

100 = πr²h

---- (1)

The Surface area cylinder is = area of the circular base + area of the circular top + area of the cylinder

S = πr² + πr² + 2πrh

S = 2 πr² + 2πrh

[substituting (1) in the volume formula]

------ (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with r and then equating it to zero. This is because if the function f(r) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to r:

[Since and ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

---- (4)

Now to check if this critical point will determine the minimum surface area of the box, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with r:

----- (5)

[Since and ]

Now let us find the value of

As , so the function S is minimum at

As S is minimum from equation (4)

V = 2πr³

Now in equation (1) we have,

h = 2r = diameter

Therefore when the total surface area of a cone is minimum, then height of the cone is equal to twice the radius or equal to its diameter.

Let r be the radius of the base and h the height of a cylinder.

The surface area is given by,

S = 2 π r² + 2 π rh

………(1)

Let V be the volume of the cylinder.

Therefore, V = πr²h

…….Using equation 1

Differentiating both sides w.r.t r, we get,

……….(2)

For maximum or minimum, we have,

⇒

⇒ S = 6πr²

2πr² + 2πrh = 6πr²

h = 2r

Differentiating equation 2, with respect to r to check for maxima and minima, we get,

Hence, V is maximum when h = 2r or h = diameter

Given,

• Volume of the sphere.

• Volume of the cone.

• Cone is inscribed in the sphere.

• Volume of cone is maximum.

Let us consider,

• The radius of the sphere be ‘a’ units.

• Volume of the inscribed cone be ‘V’.

• Height of the inscribed cone be ‘h’.

• Radius of the base of the cone is ‘r’.

Given volume of the inscribed cone is,

Consider OD = (AD-OA) =(h-a)

Now let OC² = OD² + DC², here OC = a, OD = (h-a), DC = r,

So a² = (h-a)² + r²

r² = a² – (h²+ a² – 2ah)

r² = h (2a -h) ----- (1)

Let us consider the volume of the cone:

Now substituting (1) in the volume formula,

---- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with h and then equating it to zero. This is because if the function V(r) has a maximum/minimum at a point c then V’(c) = 0.

Differentiating the equation (2) with respect to h:

[Since ]

------- (3)

To find the critical point, we need to equate equation (3) to zero.

4πah – 3πh² = 0

h(4πa-3πh)= 0

h = 0 (or)

[as h cannot be zero]

Now to check if this critical point will determine the maximum volume of the inscribed cone, we need to check with second differential which needs to be negative.

Consider differentiating the equation (3) with h:

----- (4)

[Since ]

Now let us find the value of

As , so the function V is maximum at

Substituting h in equation (1)

As V is maximum, substituting h and r in the volume formula:

Therefore when the volume of a inscribed cone is maximum, then it is equal to times of the volume of the sphere in which it is inscribed.

Given,

The pth power of a number exceeds by a fraction to be the greatest.

Let us consider,

• ‘x’ be the required fraction.

• The greatest number will be y = x - x^p ------ (1)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function y(x)has a maximum/minimum at a point c then y’(c) = 0.

Differentiating the equation (1) with respect to x:

---- (2)

[Since ]

To find the critical point, we need to equate equation (2) to zero.

1 = px^p-1

Now to check if this critical point will determine the if the number is the greatest, we need to check with second differential which needs to be negative.

Consider differentiating the equation (2) with x:

----- (3)

[Since ]

Now let us find the value of

As , so the number y is greatest at

Hence, the y is the greatest number and exceeds by a fraction

Given,

• A point is present on a curve y² = 4x

• The point is near to the point (2,-8)

Let us consider,

• The co-ordinates of the point be P(x,y)

• As the point P is on the curve, then y² = 4x

• The distance between the points is given by,

D² = (x-2)² +(y+8)²

D² = x²-4x+4 + y² + 64 + 16y

Substituting x in the distance equation