RD Sharma Class 12 Ex 18.5 Solutions Chapter 18 Maxima and Minima

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter18
Exercise18.5
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 18.5 Solutions Chapter 18 Maxima and Minima

Question 1. Determine two positive numbers whose sum is 15 and the sum of whose squares is minimum.

Solution:

Let us assume the two positive numbers are x and y, 

And it is given that x + y = 15         …..(i)

So, let P = x2 + y2                     …..(ii)

From eq (i) and (ii), we get

P = x2 + (15 – x)2

On differentiating w.r.t. x, we get

dP/dx = 2x + 2(15 – x)(-1)

= 2x -30 +2x

= 4x -30

For maxima and minima.

Put dP/dx = 0

⇒ 4x – 30 = 0

⇒ x = 15/2

Since, d2P/dx2 = 4 > 0

So, x = 15/2 is the point of local minima,

From eq(i), we get

 y = 15 – 15/2 = 15/2

So, the two positive numbers are 15/2, 15/2.

Question 2. Divide 64 into two parts such that the sum of the cubes of two parts is minimum.

Solution:

Let us assume 64 is divide into two parts that is x and y 

So, x + y = 64                    …..(i)

Let P = x3 + y3               ………(ii)

From eq(i) and (ii), we get

P = x3 + (64 – x)3

On differentiating w.r.t. x, we get

dP/dx = 3x2 + 3(64 – x)× (-1)

= 3x2 – 3(4096 – 128x + x2)

= -3 (4096 – 128x)

For maxima and minima.

Put dP/dx = 0 

⇒ -3(4096 – 128x) = 0

⇒ x = 32

Now,

d2s/dx2 = 384 > 0

So, x=32 is the point of local maxima.

Hence, the 64 is divide into two equal parts that is (32, 32)

Question 3. How should we choose two numbers, each greater than or equal to -2, whose sum is 1/2 so that the sum of the first and the cube of the second is minimum?

Solution:

Let us assume x and y be the two numbers, such that x, y ≥ -2 and

x + y = 1/2             ……(i)

So, let P = x + y3             …….(ii)

From eq(i) and (ii), we get

P = x + (1/2 – x)3

On differentiating w.r.t. x, we get

dP/dx = 1 + 3(1/2 – x)2 × (-1)

= 1 – 3(1/4 – x + x2)

= 1/4 +3x -3x2

For maximum and minimum,

Put dP/dx = 0

⇒ 1/4 + 3x – 3x2 = 0

⇒ 1 + 12x – 12x2 = 0

⇒ 12x2 – 12x – 1 = 0

⇒ x=\frac{12±\sqrt{144+48}}{24}

⇒ x = 1/2 ± (8√3/24)

⇒ x = 1/2 ± (1/√3)

⇒ x = {1/2 – (1/√3)}, {1/2 + (1/√3)}

Now,

d2P/dx2 = 3 – 6x

So, at x =1/2 – (1/√3), d2P/dx2 = 3(1 – 2(1/2 – 1/√3))

= 3(+2/√3) = 2√3 > 0

Hence, x = 1/2 – 1/√3 is point of local minima

From eq(i), we get

y = 1/2 – (1/2 – 1/√3) = 1/√3

So, the numbers are (1/2 – 1/√3) and 1/√3

Question 4. Divide 15 into two parts such that the square of one multiplied with the cube of the other minimum.

Solution:

Let us assume 15 is divide into two parts that is x and y 

So, x + y = 15

Also, P = x2y3

From eq(i) and (ii), we get

P = x2(15 – x)3

On differentiating w.r.t. x, we get

dP/dx = 2x(15 – x)3 – 3x2(15 – x)2

= (15 – x)2[30x – 2x– 3x2]

= 5x(15 – x)2(6 – x)

For maxima and minima,

Put dP/dx = 0

⇒ 15(15 – x)2(6 – x) = 0

⇒ x = 0, 15, 6

Now,

So, d2P/dx2 = 5(15 – x)2(6 – x) – 5x × 2(15 – x)(6 – x) –  5x(15 – x)2

At x = 0, d2P/dx2 = 1125 > 0

So, x = 0 is point of local minima

At x = 15, d2P/dx2 = 0

So, x = 15 is an inflection point.

At x = 6, d2P/dx2 = -2430 < 0

So, x = 6 is the point of local maxima

So, the 15 is divide into two parts that are 6 and 9.

Question 5. Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3 which has the minimum surface area?

Solution:

Let us assume r be the radius of the cylinder and h be the height of the cylinder

So, the volume of the cylinder is 100 cm3

i.e., V = πr2h = 100

h = 100/πr2……(i)

Now we find the surface area of the cylinder is

A = 2πr2 + 2πrh 

= 2πr2 + 200/r

On differentiating w.r.t. r, we get

dA/dr = 4πr – 200/r2

⇒ d2A/dr2 = 4π + 400/r3

For maxima and minima,

dA/dr = 0 

⇒ 4πr = 200/r2

⇒ r3 = 200/4π = 50/π

r = (50/π)1/3

So, when r = (50/π)1/3, d2s/dr2 > 0

Hence, from the second derivative test, the surface area is the minimum 

when the radius of the cylinder is (50/π)1/3 cm

Now put the value of r in eq(i), we get

h = 100 / π(50/π)1/3 = (2×50)/(502/3π1-2/3) = 2(50/π)1/3

Question 6. A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by

(1) M = (WL/2)x – (w/2)x2

(ii) M = Wx/3 –  (W/3) (x3/L2)

Find the point at which M is maximum in each case.

Solution:

(i) M = (WL/2)x – (w/2)x2

On differentiating w.r.t. x, we get

dM/dx = WL/2 – Wx

For maxima and minima,

Put dM/dx = 0 

WL/2 – Wx = 0   

x = L/2

Now, d2M/dx2 = -W < 0

So, x = L/2 is point of local maxima.

Hence, M is maximum when x = L/2

(ii) M = Wx/3 – (W/3)(x3/L2)

On differentiating w.r.t. x, we get

dM/dx = W/3 – Wx2/L2

For maxima and minima,

Put dM/dx = 0 

W/3 – Wx2/L2 = 0  

x = ± L/√3

Now, d2M/dx2 = – 2xW/L2

So, at x = L/√3, ⇒ d2M/dx2 =-2W/√3L < 0 (for max value)

at x = -L/√3, ⇒ d2M/dx2 = 2W/√3L > 0 (for min value)

Hence, M is maximum when x = L/√3

Question 7. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made square and the other into a circle. What should be the lengths of the two pieces so that combined area of the circle and the square is minimum?  

Solution:

Let us assume l m be the piece of length cut from the given wire to make a square.

and  the other piece of wire that is used to create a circle is of length (28-l) m.

So, the side of square = l/4

Now, let us considered the radius of the circle is r. 

Then, 2πr = 28 – l ⇒ r = (1/2π)(28 – l)

Now we find the combined area of square and circle

A = l2/16 + π[(1/2π)(28 – l)]2

= l2/16 + 1/4π (28 – l)2

On differentiating w.r.t. l, we get

dA/dl = 2l/16 + (2/4π)(28 – l)(-1)

= l/8 – (1/2π)(28 – l)

d2A/dl2 = l/8 + (1/2π) > 0

For maxima and minima,

Put dA/dl = 0 

⇒ l/8 – (1/2π)(28 – l) = 0

⇒ {πl – 4(28 – l)}8π = 0

⇒ (π + 4)l – 112 = 0

⇒ l = 112/(π + 4)

So, at l = 112/(π + 4), d2A/dl2 > 0

Hence, using second derivative test, the area is the minimum when l = 112/(π + 4)

So, the length of the two pieces of wire are 112/(π + 4) and 28π/(π + 4) cm.

Question 8. A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the wire should be cut so that the sum of the areas of the square and triangle is minimum?  

Solution:

According to the question

The length of the wire is 20 m

and the wire cut into two pieces x and y. So, x length wire is used to make a square and

y length wire is used to make triangle.

Now.

x + y = 20           …..(i)

x = 4l and y = 3a

So, A = sum of area of square and triangle

A = l+ √3/4a2         ……(ii)

We have, 4l + 3a  =20

4l = 20 – 3a

l = (20 – 3a) / 4

From eq(i), we have,

A = (20 – 3a)2/4 + √3/4a2

On differentiating w.r.t. a, we get

dA/da = 2{(20 – 3a)/4}(-3/4) + 2a × √3/4

For maxima and minima,

Put dA/da = 0

⇒ 2 {(20 – 3a)/4}(-3/4) + 2a × √3/4 = 0

⇒ -3(20 – 3a) + 4a√3 = 0

⇒ -60 + 9a + 4a√3 = 0

⇒ 9a + 4a√3 = 60

⇒ a(9 + 4√3) = 60

⇒ a = 60/(9 + 4√3)

Again differentiating w.r.t. a, we get

d2s/da2 = (9 + 4√3)/8 > 0

So, the sum of the areas of the square and triangle is minimum when a = 60 / (9 + 4√3)

So l = (20 – 3a)/4

⇒ l = \frac{20-3(60/(9+4√3))}{4}

⇒ l = (180 + 80√3 – 180)/{4(9 + 4√3)}

⇒ l = 20√3/(9 + 4√3)

Question 9. Given the sum of the perimeters of a square and a circle show that the sum of their areas is least when one side of the square is equal to the diameter of the circle.

Solution:

Let us assume the radius of the circle is r 

We have,

2πr + 4a = k (k is constant)

a = (k – 2πr)/4

Now we find the sum of the areas of the circle and the square:

A = πr2 + a2 = πr2 + (k – 2πr)2/16

On differentiating w.r.t. r, we get

dA/dr = 2πr + 2(k – 2πr)(-2π)/16 

= 2πr- π(k – 2πr)/4

For maxima and minima,

Put, dA/dr = 0

2πr = π(k – 2πr)/4

8r = k – 2πr

r = k / (8 + 2π)= k / 2(4 + π)

Now, d2A/dr2 = 2π + π2/2 > 0

So, at r = k / 2(4 + π), d2A/dr2 > 0

Hence, the sum of the areas minimum when r = k / 2(4 + π)

So, a = \frac{k-2π[\frac{k}{2}(4 + π)]}{4}

= 2r 

Hence Proved

Question 10. Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.

Solution:

Let us assume PQR is a right-angled triangle, 

So, the hypotenuse h = PR = 5 cm.

Now, le us assume a and b be the remaining sides of the triangle. 

So, a2 + b2 = 25         ……(i)

Now we find the area of PQR = 1/2 QR × PQ

A = 1/2 ab             ……(ii)

From eq(i) and (ii), we get

⇒ A = 1/2 x √(25 – a2)

On differentiating w.r.t. a, we get

dA/da = \frac{1}{2}[\frac{\sqrt{25-a^2}-2a^2}{2\sqrt{25-a^2}}]

\frac{1}{2}[\frac{25-a^2-a^2}{\sqrt{25-a^2}}]

\frac{1}{2}[\frac{25-2a^2}{\sqrt{25-a^2}}]

For maxima and minima,

Put dA/da = 0

⇒ \frac{1}{2}[\frac{25-2a^2}{\sqrt{25-a^2}}] =0

⇒ a = 5/√2

Now,

d2A/d2a = \frac{1}{2}[\frac{{\sqrt{25-a^2} (-4a)+\frac{(25-2a^2)2a}{2\sqrt{25-a^2}}}}{25-a^2}]

At a = 5√2, d2s/d2 = \frac{1}{2}[\frac{\frac{-25}{√2}×5√2+0}{\frac{25}{2}}]

= – 5/2 < 0

So, x = 5/√2 is a point local maxima,

Hence, the largest possible area of the triangle

= 1/2 × (5/√2) × (5/√2) = 25/4 square units

Question 11. Two sides of a triangle have lengths ‘a’ and ‘b’ and the angle between them is θ. What value of θ will maximize the area of the triangle? Find the maximum area of the triangle also.

Solution:

Let us assume ABC is a triangle such that AB = a, BC = b and ∠ABC = θ

and AD in perpendicular to BC.

BD = asinθ

So, the area of △ABC = 1/2 × BC × AD

⇒ A = 1/2 × b × a × sinθ

On differentiating w.r.t. θ, we get

dA/dθ = 1/2 × abcosθ

For maxima and minima,

Put dA/dθ = 0

⇒ 1/2 × abcosθ = 0

⇒ cosθ = 0

⇒ θ = π/2

Now, d2A/dθ2 = -1/2 ab sinθ

At θ = π/2, d2A/dθ= -1/2ab < 0

So, θ = π/2, is point of local maxima

Hence, the maximum area of the ABC triangle is 1/2 × absin(π/2) = 1/2 ab.

Question 12. A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each comer and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Also, find this maximum volume.  

Solution:

Let us assume that x cm be the side of the square to be cut off.

Now, the length and the breadth of the box will be (18 – 2x) cm each and 

the x cm be the height of the box.

So, the volume of the box is 

V (x) = x(18 – 2x)2

On differentiating w.r.t. x, we get

V'(x) = (18 – 2x)2 – 4x(18-2x)

= (18 – 2x)[18 – 2x – 4x]

= (18 – 2x)(18 – 6x)

= 6 × 2(9 – x)(3 – x)

= 12(9 – x)(3 – x)

Again on differentiating w.r.t. x, we get

V”(x) = 12 [-(9 – x) – (3 – x)]

= -12 (9 – x + 3 – x)

= -12 (12 – 2x)

= -24 (6 – x)

For maxima and minima,

Put V'(x) = 0

12(9 – x)(3 – x) = 0

x = 9, 3

When x = 9 length and breadth of the box become zero.

So, x ≠ 9

When x = 3, V”(x) = -24 (6 – x) = -72 < 0

So, x = 3 is the point of maxima

Hence, the maximum volume is Vx = 3 = 3(18 – 2 × 3)2

⇒ V = 3 × 122

⇒ V = 3 × 144

⇒ V = 432 cm3 

Question 13. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

Solution:

Let us assume that x cm be the side of the square to be cut off.

So, the height of the box = x, 

the length of the box = 45 – 2x, 

and the breath of the box = 24 – 2x.

So, the volume of the box is 

V(x) = x (45 – 2x)(24 – 2x)

= x (1080 – 90x – 48x + 4x2)

= 4x3 – 138x2 + 1080x

On differentiating w.r.t. x, we get

V ‘(x)= 12x2 – 276x + 1080

= 12(x– 23x + 90)

= 12(x – 18) (x – 5)

Again on differentiating w.r.t. x, we get

V ”(x) = 24x – 276 = 12 (2x – 23)

For maxima and minima,

Put V'(x) = 0

4x3 – 138x2 + 1080x = 0

x(x – 18) – 5(x – 18) = 0

(x – 5)(x – 18) = 0

So, x = 18 and x = 5

when x = 18 it is not possible to cut off a square of side 18 cm from each corner of the rectangular sheet. 

So, x ≠ 18

When x = 5, V ”(5) = 12 (10 – 23) = 12(-13) = -156 < 0

So, x = 5 is the point of maxima.

Hence, the volume of the box is maximum when x = 5.

Question 14. A tank with rectangular base and rectangular sides open at the top is to be constructed so that its depth is 2 m and volume is 8m3. If building of tank costs ₹ 70 per square meter for the base and ₹ 45 per square meter for sides, what is the cost of least expensive tank?

Solution:

Let us considered the length, breadth and height of the tank be l, b, and h

According to the question

The height of the tank is 2 and the volume is 8m3

So, the volume of the tank is  

V = l × b × h 

8 = l × b × 2

lb = 4  

⇒ b = 4/l    ….(i)

Now, we find the area of the base = lb = 4

and the area of the four walls (A) = 2h (l + b)

A = 4 (l + l/4)

On differentiating w.r.t. l, we get

⇒ dA/dl = 4 (l – 4/l2)

Again differentiating w.r.t. l, we get

 d2A/dl2 = 32/l3

For maxima and minima,

Put dA/dl = 0

⇒ l – 4/l2 = 0 

⇒ l= 4

⇒ l = ±2

As we know that the length cannot be negative. So, l ≠ 2

When l = 2, d2A/dl= 32/8 = 4 > 0

So, l = 2 is the point of minima.

Now put the value of l = 2 cm in the eq(i)

b = 4/l = 4/2 = 2

So, l = b = h = 2

Hence, the area is the minimum when l = 2.

So, the cost of building the base = 70 × (lb) = 70 × 4 = 280

Cost of building the walls = 2h (l + b) × 45 = 90 × 2 × (2 + 2) = 8 × 90 = 720

Hence, the total cost = 280 + 720 = 1000

Question 15. A window in the form of a rectangle is surmounted by a semi-circular opening. The total perimeter of the window is 10 m. Find the dimensions of the rectangular part of the window to admit maximum light through the whole opening.

Solution:

Let us assume x and y be the length and breadth of the rectangle. 

So the radius of the semicircular opening = x/2

From the question it is given that the perimeter of the window is 10 m.

So, x + 2y + πx/2 = 10

⇒ x(1 + π/2) + 2y = 10

⇒ 2y = 10 – x(1 + π/2)

⇒ y = 5 – x(1/2 + π/4)

Now, the area of the window is

A = xy + (π/2)(x/2)2

= x [5 – x(1/2 + π/4)] + (π/8)x2

= 5x – x2(1/2 + π/4) + (π/8)x2

On differentiating w.r.t. x, we get

dA/dx = 5 – 2x(1/2 + π/4) + (π/4)x

= 5 – x(1 + π/2) + (π/4)x

Again differentiating w.r.t. x, we get

d2A/dx2 = -(1 + π/2) + π/4 = -1 – π/4

For maxima and minima,

Put dA/dx = 0

⇒ 5 – x(1 + π/2) + (π/4)x = 0

⇒ 5 – x – (π/4)x = 0

⇒ x (1 + π/4) = 5

⇒ x = 5/(1 + π/4) = 20/(π + 4)

So, when x = 20/(π + 4), d2A/dx2 < 0

So, the area is the maximum when length (x) = 20/(π + 4)

Now,

y = 5 – 20/(π + 4){(2 + π)/4} = 5 – 5(2 + π)/(π + 4) = 10/(π + 4) m 

Hence, the length of the rectangle is 20/(π + 4) m and the breadth is 10/(π + 4) m.

Question 31. Find the point on the parabola x= 2y which is closest to the point (0, 5).

Solution:

The given equation of the curve is 

x= 2y ……(i)

Let P(x, y) be a point on the given curve, and 

Q be the square of the distance between P and A(0, 5). 

Q = x² + (y – 5)2  ……..(ii)

= 2y + (y – 5)2

On differentiating w.r.t. y, we get

dQ/dy = 2 + 2(y – 5)

For maxima and minima,

Put dQ/dy = 0

⇒ 2 + 2y – 10 = 0

⇒ y = 4

Now,

When y = 4, d2Q/dy2 = 2 > 0

So, y = 4 is the point of local minima

Now put the value of y in eq(1), we get

x = ±2√2

So, P(±2√2, 4) is the closest point on the curve to A(0, 5).

Question 32 Find the coordinates of a point on the parabola y = x+ 7x + 2 which is closest to the straight line y = 3x – 3.

Solution:

The given equation of parabola is

y = x2 + 7x + 2   ……(i)

closest to the straight line y = 3x – 3         ……(ii)

Let us considered P(x, y) be the point on the given parabola which is closest to the line y = 3x – 3

Let Q be the perpendicular distance from P to the line y = 3x – 3

Q = \frac{|y - 3x+3|}{\sqrt{1^2+(-3)^2}}

\frac{|x² + 7x + 2-3x +3|}{√10}

On differentiating w.r.t. x, we get

dQ/dx = (2x +4) / √10

For maxima or minima, we have

Put dQ/dx = 0

⇒ (2x + y)/√10 = 0

⇒ x = -2

Now put the value of x in eq(i), we get

 y = 4 – 14 + 2 = -8

When x = -2 and y = -8, d2Q/dx2 = 2/√10 > 0

So, x = -2, and y = -8 is the point of local minima,

Hence, P(-2, -8)is the closest point on the parabola to the line y = 3x-3.

Question 33 Find the point on the curve y= 2x which is at a minimum distance from the point (1, 4).

Solution:

The given equation of the curve is 

y= 2x  …..(i)

Let P(x, y) be a point on the given curve, which is minimum distance from the point A(1, 4) and 

square of the length of AP

Q = (x – 1)+ (y – 4)2

= x2 + 1 – 2x +y2 +16 – 8y

= x– 2x +2x+17 – 8y

= y4/4 – 8y +17              [Since y= 2x]

On differentiating w.r.t. y, we get

dQ/dy = y3 – 8

For maxima and minima, we have

Put dQ/dy = 0

y3 – 8 = 0

y3 = 23

y = 2

When y = 2, d2S/dy2 = 3y= 12 > 0

So, y = 2 is the point of local minima,

We have

x = y2/2 = 4/2 = 2

Hence, P(2, 2) is at a minimum distance from the point A(1, 4).

Question 34. Find the maximum slope of the curve y = -x+ 3x+ 2x – 27.

Solution:

The given equation of curve is

y = x3 + 3x2 + 2x – 27          …..(i)

The slope of the given equation of the curve is 

m = dy/dx = −3x2 + 6x + 2     …..(ii)

On differentiating w.r.t. x, we get

dm/dx = -6x + 6

Again differentiating w.r.t. x, we get

d2m/dx2 = -6 < 0

For maxima and minima,

Put dm/dx = 0

⇒ -6x + 6 = 0

⇒ x = 1

When x = 1, d2m/dx= -6 < 0

So, x = 1 is point of local maxima

Hence, the maximum slope of the given curve is  = -3 + 6 + 2 = 5

Question 35. The total cost of producing x radio sets per day is (x2/4 + 35x + 25) and the price per set at which they may be sold is (50 – x/2). Find the daily output to maximize the total profit. 

Solution:

Given, 

The cost of producing x radio sets is Rs. x2/4 + 35x + 25

And selling price of x radio is Rs. x(50 – x/2)

So, 

The profit on x radio sets is

P = 50x – x2/2 – x2/4 -35x – 25

On differentiating w.r.t. y, we get

dp/dx = 50 – x – x/2- 35

= 15 – 3x/2

For maxima and minima,

Put, dp/dx = 0

⇒ 15 – 3x/2 = 0

⇒ x = 10

When x = 10, d2p/dx2 = -3/2 <0

So, x = 10 is the point of local maxima

Hence, the profit is maximum only if the daily output is 10.

Question 36. Manufacturers can sell x items at a price of (5 – x/100) each. The cost price is (x/5 + 500). Find the number of items he should sell to earn maximum profit.  

Solution:

Let us considered S(x) be the selling price of x items and C(x) be the cost price of x items.

So, it is given that

S(x) = (5 – x/100) = 5x – x2/100

and            

C(x)= x/5 + 500

Then the profit function is 

P(x) = S(x) – C(x) 

= 5x – x2/100 -x/5 – 500 = 24x/5 – x2/100 – 500

On differentiating w.r.t. x, we get

P'(x) = 24/5 – x/50

Also p”(x) = – 1/50

For maxima and minima,

Put, P'(x) = 0

⇒ 24/5 – x/50 = 0

⇒ x = 24/5 × 50 = 240

When x = 240, P”(240) = -1/50 < 0

So, x = 240 is a point of maxima.

Hence, when the manufacturer sells 240 then he can earn maximum profit.

Question 37. An open tank is to be constructed with a square base and vertical sides so as to contain a given quantity of water. Show that the expenses of lining with lead will be least, if depth is made half of width.

Solution:

Let us considered l be the length of side of square base of the tank and h be the height of tank.

So, the volume of tank is 

V = l2h  ……(i)

And the total surface area is 

A = l2 + 4lh  …..(ii)

Now,

From eq(i) and (ii), we get

A = l2 + 4v/l

On differentiating w.r.t. l, we get

dA/dl = 2l – 4v/l2

Also, d2A/dl2 = 2 + 8v/l3

For maxima and minima,

Put, dA/dl = 0

⇒ 2l – 4v/l2 = 0

⇒ 2l3 – 4v = 0

⇒ l3 = 2v = 2t2h

⇒ l2[ l – 2h ]= 0

⇒ l = 0 or 2h

Here, l = 0 is not possible

At l = 2h, d2S/dl2 > 0      

So, l = 2h is point of local minima

Hence, the total surface area is minimum when l = 2h

Question 38. A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square meter as that in the bottom. What are the most economic dimensions?

Solution:

Let us considered ABCDEFGH be a box of constant volume c and it is given that the box is twice as long as its width.

So, BF = x and AB = 2x

So, the cost of material of top and front side = 3 x cost of material of the bottom of the box.

⇒ 2x × x + xh + xh+ 2xh + 2xh = 3 × 2x2

⇒ 2x2 + 2xh + 4xh = 6x2

⇒ 4x2 – 6xh = 0

⇒ 2x(2x – 3h) = 0

⇒ x = 3h/2 or h = 2x/3

Volume of box is 

V = 2x × x × h

c = 2x2h

h = c/2x2                           ……(i)

Now,

The surface are of the box is 

A = 2 (2x+ 2xh + xh)

= 2(2x2 + 3xh)

= 2(2x2 + 3xc/2x2)

= 2(2x2 + 3/2 × c/x)

On differentiating w.r.t. x, we get

dA/dx = 2(4x – 3/2 × c/x2)

For maxima and minima,

Put dA/dx = 0

2(4x – 3/2 × c/x2) = 0

8x3 – 3c = 0

x = (3c/8)1/3

When x = (3c/8)1/3, d2A/dx2 = 2 (4 + 3 × C/x3) > 0  

So, x = (3c/8)1/3 is point of local minima

Hence, the most economic dimension will be

x = width = (3c/8)1/3

2x = length = 2(3c/8)1/3

h = height = 2x/3 = 2/3 × (3c/8)1/3

Question 39. The sum of the surface areas of a sphere and a cube is given Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

Solution:

Lets us considered S be the sum of the surface areas of a sphere and a cube.

S = 4πr2 + 6l2       …..(i)

Here, I be the side of the cube and r be the radius of the sphere

And V be the volume of sphere and cube

V = 4/3πr3 + l3       …..(ii)

I=\sqrt{\frac{S-4πr^2}{6}}
V=4/3πr^2+(\frac{S-4πr^2}{6})^\frac{3}{2}

On differentiating w.r.t. r, we get

\frac{dV}{dr}=4πr^2+\frac{3}{2}(\frac{s-4πr^2}{6})^\frac{1}{2} (\frac{-4π}{6})^{2r}

For maxima and minima,

Put dV/dr = 0

⇒ 4πr= π/6(S – 4πr2)1/2 × 2r = 0

⇒ 2rπ[2r – I] = 0

r = 0, l/2

Now,

\frac{d^2V}{dr^2}=8πr-\frac{2π}{√6}[(S-4πr^2)^{\frac{1}{2}}]-\frac{8πr^2}{2(S-4πr^2)^{\frac{1}{2}}}

At r = I/2, d2V/dr2 > 0

So, r = I/2 is point of local minima

Hence, the volume is minimum when l = 2r

Question 40. A given quantity of metal is to be cast into a half-cylinder with a rectangular base and semi-circular ends. Show that in order that the total surface area may be minimum, the ratio of the length of the cylinder to the diameter of its semi-circular ends is π:(π + 2)

Solution:

Let ABCDEF be a half cylinder with rectangular base and semicircular ends.

So, AB = height of the cylinder = h

Let us considered r be the radius of the cylinder.

As we know that the volume of the half cylinder is 

V = 1/2πr2h

2V/πr2 = h

The total surface area of the half cylinder is

A = LSA of the half cylinder + area of two semicircular ends + area of the rectangle (base)

A = πrh + πr2/2 + πr2/2 +h2r

= (πr + 2r)h + πr2

= (π + 2)2v/πr + πr2

On differentiating w.r.t. r, we get

dA/dr = [(π + 2)2v/π(-1/r2) + 2πr]

For maxima and minima,

Put dA/dr = 0 

⇒ [(π + 2)2v/π(-1/r2) + 2πr] = 0 

⇒ [(π+2)2v/πr= 2πr

But 2r = D

h : D = π : π + 2

Again differentiating w.r.t. r, we get

d2A/dr2 = (π + 2)v/π × 2/r+ 2π > 0

So, S will be minimum when h : 2r is π : π-12.

Hence, the height of the cylinder : Diameter of the circular end π : π + 2

Question 41. The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a?

Solution:

Let us considered ABCD be the cross-sectional area of the beam which is cut from a circular log of radius a.

So, AO = a and AC = 2a

Let x be the width and y be the depth of log. Also, S be the strength of the beam 

According to the question,

S = xy2      …..(i)

In ΔABC

x2 + y2 = (2a)2

⇒ y = (2a)2 – x2       …..(ii)

From eq(i) and (ii), we get

S = x((2a)2 – x2)

On differentiating w.r.t. x, we get

⇒ dS/dx = (4a2 – x2) – 2x2

⇒ dS/dx = 4a2 – 3x2

For maxima or minima

Put, dS/dx = 0

⇒ 4a2 – 3x2 = 0

⇒ x2 = 4a2/3

x = 2a√3

Now put the value of x in eq(ii)

y2 = 4a2 – 4a2/3 = 8a2/3

y = 2a×√(2/3)

Now, 

At x = 2a/√3, y = √(2/3)2a, d2S/dx2 = -6x = -12a/√3 < 0

So, (x = 2a/√3, y = √(2/3)2a) is the point of local maxima.

Hence, the dimension of the strongest beam is 2a/√3 and √(2/3)2a.

Question 42. A straight line is drawn through a given point P(1, 4). Determine the least value of the sum of the intercepts on the coordinate axes.

Solution:

Let us considered l be a line through the point P (1, 4) that cuts the x-axis and y-axis.

So, the equation of line(l) is 

y – 4 = m(x – 1)

x-Intercept is (m – 4)/m and y-intercept is (4-m)

Let S = (m – 4)/m + 4 – m

On differentiating w.r.t. m, we get

dS/dm = 4/m2 – 1

For maxima and minima,

Put, dS/dm = 0

⇒ 4/m2 -1 = 0

⇒ m = ±12

Now,

d2S/dm2 = -8/m3

At m = 2, d2S/dm= -1 < 0

At m = -2, d2S/dm2= 1 > 0

So, m = -2 is point of local minima.

Hence, the least value of sum of intercept is

= (m – 4)/m + 4 – m

= 3 + 6 = 9

Question 43. The total area of a page is 150 cm2. The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. What must be the dimensions of the page in order that the area the printed matter may be maximum?

Solution:

Given that the area of the page PQRS in 150 cm2

Also, AB + CD = 3 cm

EF + GH = 2 cm

Let us considered x and y be the combined width of margin at the top and bottom and the sides

x = 3 cm and y = 2 cm.

Now, we find the area of printed matter = area of P’Q’R’S’

⇒ A = P’Q’ Q’R’

⇒ A = (b – y)(l – x)

⇒ A = (b – 2)(l – 3)          …..(i)

Also,

The area of PQRS = 150 cm2

⇒ lb = 150                  …..(ii)

From eq(i) and (ii), we get

A = (b – 2)(150/b – 3)

On differentiating w.r.t. b, we get

dA/db = (150/b – 3) + (b – 2)(- 150/b2)

For maximum and minimum,

Put dA/db = 0

⇒ (150 – 3b)/b + (-150)(b – 2)b2 = 0

⇒ 150b – 3b2 – 150b +300 = 0

⇒ -3b2 + 300 = 0

⇒ b = 10

From eq(ii), we get

l = 15

Now,

d2A/db2 = -150/b2 – 150[-1/b2 + 4/b3]

When b = 10, d2A/db2 = -15/10 – 150[-1/100 + 4/1000] = -1.5 + 9 = -0.6 < 0

So, b = 10 is point of local maxima.

Hence, the required dimension will be 15 cm and 10 cm.

Question 44. The space s described in time t by a particle moving in a straight line is given by s = t5 – 40t3 + 30t2 + 80t – 250. Find the minimum value of acceleration.

Solution:

Given that s is the space in time t by a moving particle is 

S = t5 – 40t3 +30t2 +80t – 250

Velocity = dS/dt = 5t4 -120t2 + 60t + 80

Acceleration = a = d2S/dt2 = 20t3 – 240t + 60t           …..(i)

Now,

da/dt = 60t2 – 240

For maxima and minima,

Put, da/dt = 0

⇒ 60t2 – 240 = 0

⇒ 60(t2 – 4) = 0

⇒ t = 2

Now,

d2a/dt2 = 120t

At t = 2, d2a/dt2 = 240 > 0

So, t = 2 is point of local minima

Hence, the minimum acceleration is 160 – 480 + 60 = -260 

Question 45. A particle moving in a straight line such that its distance s at any time t is given by s = t4/4 – 2t+ 4t– 7. Find when its velocity is maximum and acceleration minimum.

Solution:

Given that 

Distance(S) = t4/4 – 2t3 + 4t2 – 7

Velocity(V) = dS/dt = t3 – 6t2 + 8t

Acceleration(a) = d2S/dt2 = 3t2 – 12t + 8

For velocity to be maximum and minimum,

Put dV/dt = 0

⇒ 3t2 – 12t +8 = 0

⇒ t = \frac{12±\sqrt{144-96}}{6}

= 2 ± 4√3/6

t = (2 + 2/√3), (2 – 2/√3)

Now,

d2V/dt2 = 6t – 12

At t = (2 – 2/√3), d2V/dt2 = 6(2 – 2/√3) – 12 = -12/√3  < 0

t = (2 + 2/√3), d2V/dt2 = 6(2 + 2/√3) – 12 = 12/√3  > 0

So, at t = (2 – 2/√3), velocity is maximum

For acceleration to be maximum and minimum

Put da/dt = 0

⇒ 6t – 12 = 0

⇒ t = 2

Now,

When, t = 2, d2a/dt2 = 6 > 0

Hence, at t = 2, acceleration is minimum.

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