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ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.8

Question 1.
Find two numbers whose sum is 24 and whose product is as large as possible. (NCERT)
Solution:
Let the one number be x, (0 < x < 24)
∴ other number be 24 – x because sum of two numbers be 24.
Let P = x (24 – x) ……..(1)
Duff. (1) w.r.t. x; we have
$$\frac{d P}{d x}$$ = 24 – 2x
For maximum/minima, $$\frac{d P}{d x}$$ = 0
⇒ 24 – 2x = 0
⇒ x = 12
∴ $$\frac{d^2 P}{d x^2}$$ = – 2
⇒ ($$\frac{d^2 P}{d x^2}$$)x = 12 = – 2 < 0
Thus, x = 12 be a point of local maxima and x = 12 be the only point of local maxima.
So the local maximum value be the absolute maximum value.
Thus, required numbers are 12 and (24 – 12) i.e. 12.

Question 2.
The sum of two positive numbers is 20. Find the numbers
(i) if the sum of their squares is minimum.
(ii) if the product of the square of one and the cube of the other is maximum.
Solution:
(i) Let the required numbers be x, 20 – x,
since the sum of two positive integers is 20.
Let S = x2 + (20 – x)2
Diff. both sides w.r.t. x, we have
$$\frac{d S}{d x}$$ = 2x + 2 (20 – x) (- 1)
= 2x – 40 + 2x
= 4x – 40
For maxima/minima, $$\frac{d S}{d x}$$ = 0
⇒ 4x – 40 = 0
⇒ x = 10
∴ ($$\frac{d^2 P}{d x^2}$$)x = 10 = 4 > 0
Thus x = 10 be a point of local minima.
∴ required numbers are 10 and (20 – 10) i.e. 10 and 10.

(ii) Let the required numbers be x and 20 – x
Since the sum of two numbers be 20.
Let P = x2 (20 – x)3 ;
Diff. both sides w.r.t. x, we have
$$\frac{d P}{d x}$$ = 3x2 (20 – x)2 (- 1) + (20 – x)3 2x
= – x (20 – x)2 [3x – 2 (20 – x)]
= – x (20 – x)2 (5x – 40)
For maxima / minima, $$\frac{d P}{d x}$$ = 0
⇒ x (20 – x)2 (5x – 40) = 0
⇒ x = 0, 20, 8
But 0 < x < 20 ∴ x = 8
When x slightly < 8,
$$\frac{d P}{d x}$$ = (- ve) (+ ve) (+ ve) = – ve
∴ $$\frac{d P}{d x}$$ changes its sign from +ve to – ve.
Hence x = 8 be a point of local maxima.
Thus the required numbers are 8 and 20 – 8 i.e. 8 and 12.

Question 3.
The product of two positive numbers is 16. Find the numbers
(i) if their sum is least.
(ii) if the sum of one and the square of the other is least.
Solution:
(i) Let the two numbers be x and $$\frac{16}{x}$$,
since the product of two numbers be 16.
Let S = x + $$\frac{16}{x}$$
∴ $$\frac{d S}{d x}$$ = 1 – $$\frac{16}{x^2}$$
For maxima/minima, $$\frac{d P}{d x}$$ = 0
⇒ 1 – $$\frac{16}{x^2}$$ = 0
⇒ x2 = 16
⇒ x = ± 4 but x, y ∈ N
∴ x = 4
and $$\frac{d^2 S}{d x^2}$$ = $$\frac{32}{x^3}$$
∴ $$\left(\frac{d^2 S}{d x^2}\right)_{x=4}=\frac{32}{64}=\frac{1}{2}$$ > 0
∴ x = 4 be a point local minima.
Thus S is least for x = 4
∴ required numbers are 4 and $$\frac{16}{4}$$ i.e. 4 and 4.

(ii) Let the required two numbers be x and $$\frac{16}{x}$$,
Since the product of two numbers be 16.
Let S = x2 + 16/x
$$\frac{d S}{d x}$$ = $$\frac{-16}{x^2}$$ + 2x
∴ $$\frac{d^2 S}{d x^2}$$ = $$\frac{32}{x^3}$$ + 2
For maxima/minimum, $$\frac{d S}{d x}$$ = 0
⇒ $$\frac{-16}{x^2}$$ + 2x = 0
⇒ x3 = 8
⇒ x = 2
at x = 2;
$$\frac{d^2 S}{d x^2}$$ = $$\frac{32}{8}$$ + 2
= 4 + 2 = 6 > 0
∴ x = 2 be a point of local minima.
Thus S is minimum for x = 2.
Hence the required numbers are 2 and $$\frac{16}{2}$$ i.e. 2 and 8.

Question 4.
Find the positive numbers x and y such that x + y = 60 and xy is maximum. (NCERT)
Solution:
Given x + y = 60 …………(1)
where x, y > 0
Let P = xy3
= y3 (60 – y)
∴ $$\frac{d P}{d y}$$ = 180y2 – 4y3
= 4y2 [45 – y]
For maxima/minina, $$\frac{d P}{d y}$$ = 0
⇒ 4y2 (45 – y) = 0
⇒ y = 0, 45
but y > 0
∴ y = 45
and $$\frac{d^2 P}{d y^2}$$ = 12 × 45 × (- 15) = – 8100 < 0
∴ P is maximise for y = 45
∴ from (1) ; x = 15.
Hence the required two numbers be 15, 45.

Question 5.
Find two positive numbers x andy such that their sum is 35 and the product x2y5 is maximum. (NCERT)
Solution:
Let the two positive numbers be x and y.
Given x + y = 35 ………..(1)
Let P product = x2y5
⇒ P = (35 – y2) y5
∴ $$\frac{d P}{d y}$$ = (35 – y)2 5y4 ± y5 2(35 – y) (- 1)
= y4 (35 – y) [5 (35 – y) – 2y]
= y4 (35 – y) (175 – 7y)
For maxima/minima, $$\frac{d P}{d y}$$ = 0
⇒ y = 0, 35, 25
but y ≠ 0, 35
[∵ y > 0 and further if y = 35 then x = 0]
∴ possible value of y = 25.
When y slightly < 25 ⇒ $$\frac{d P}{d y}$$ = (+ve) (+ve) (+ve) = +ve When y slightly > 25
⇒ $$\frac{d P}{d y}$$ = (+ve) (+ve) (-ve) = -ve
∴ $$\frac{d P}{d x}$$ varies from +ve to -ve
∴ P is maximise for y = 25
∴ from (1); x = 10
∴ Required two numbers are 10, 25.

Question 6.
(i) Find two positive numbers whose sum is 16 and the sum of whose cube is minimum.
Solution:
(i) Let the two numbers be x andy s.t. x, y > 0
also given, x + y = 16
Let S = x<sup3 + y3
= x3 + (16 – x)3
∴ $$\frac{d S}{d x}$$ = 3x2 + 3 (16 – x)2 (- 1)
∴ $$\frac{d^2 S}{d x^2}$$ = 3 [2x + 2 (16 – x)] = 96 > 0
For maximaJminima, $$\frac{d S}{d x}$$ = 0
⇒ 3 [x2 – (16 – x)2] = 0
⇒ x2 – (256 + x2 – 32x) = 0
⇒ x = $$\frac{256}{32}$$ = 8
∴ ($$\frac{d^2 S}{d x^2}$$)x = 8 = 3 (16 + 16) = 96 > 0
∴ S is minimum for x = 8 and
from (1) ; y = 8
Thus the required two numbers are 8 and 8.

Question 6 (old).
(i) Find two positive numbers whose sum is 15 and the sum of whose squares is minimum. (NCERT)
Solution:
Let the two numbers be x and y s.t. x, y > 0
Given x + y = 15
Let S = x2 + y2
= x2 + (15 – x)2 …………….(2)
∴ $$\frac{d S}{d x}$$ = 2x + 2 (15 – x) (- 1)
= 2 [x – 15 + x]
= 2 (2x – 15);
$$\frac{d^2 S}{d x^2}$$ = 4
For maxima/minima
∴ $$\frac{d S}{d x}$$ = 0
⇒ x = $$\frac{15}{2}$$
∴ $$\frac{d^2 S}{d x^2}$$ = 4 > 0
Thus S is minimise for x = $$\frac{15}{2}$$
∴ from (1) ;
y = 15 – $$\frac{15}{2}$$ = $$\frac{15}{2}$$
Thus required two positive numbers are $$\frac{15}{2}$$ and $$\frac{15}{2}$$.

Question 7.
Three numbers are given whose sum is 180 and the ratio of first two of them is 1 : 2. If the product of the numbers is greatest, find the numbers. (ISC 2003)
Solution:
Let the first two numbers be x and 2x, since the ratio of first two of them be 1 : 2.
Also the s um of first three numbers be 180. required numbers are x, 2x, 180 – 3x.
Let P = x × 2x × (180 – 3x)
= 2x2 (180 – 3x)
= 6 (60x2 – x3)
Diff. both sides w.r.t. x, we have
$$\frac{d P}{d x}$$ = 6 [120x – 3x2] ;
$$\frac{d^2 P}{d x^2}$$ = 6 (120 – 6x)
For maxima/minima, $$\frac{d P}{d x}$$ = 0
⇒ 120x – 3x2 = 0
⇒ – 3x [x – 40] = 0
⇒ x = 0, 40
But x > 0
∴ x = 40
∴ ($$\frac{d^2 P}{d x^2}$$)x = 40 = 6 (120 – 240)
= – 720 < 0
∴ x = 40 be a point of maxima.
Thus product P is maximum for x = 40.
Hence the required numbers are, 40, 2 × 40 and 180 – 120 i.e. 40, 80 and 60.

Question 8.
Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 + 72x – 18x2. (NCERT)
Solution:
Given profit function p(x) = 41 + 72x – 18x2
Diff. both sides w.r.t. x, we have
$$\frac{d P}{d x}$$ = 72 – 36x
For maxima/minima, $$\frac{d P}{d x}$$ = 0
⇒ 72 – 36x = 0
⇒ x = 2
Also, $$\frac{d^2 y}{d x^2}$$ = – 36
∴ ($$\frac{d^2 P}{d x^2}$$)x = 2 = – 36 < 0
Thus x = 2 be a point of maxima.
∴ profit P(x) will be maximum when x = 2.
and Maximum profit = p (2) = 41 + 144 – 72 = 113 units.

Question 9.
The cost (in Rs.) per article C, of manufacturing a certain article is given by the formula C = 5 + $$\frac{48}{x}$$ + 3x2, where x is the number of articles manufactured per hour. Find the minimum value of C.
Solution:
Given C = 5 + $$\frac{48}{x}$$ + 3x2
Diff. both sides w.r.t. x, we have
$$\frac{d C}{d x}$$ = – $$\frac{48}{x^2}$$ + 6x
∴ $$\frac{d^2 C}{d x^2}$$ = $$\frac{96}{x^3}$$ + 6
For maxima/minima, we have
$$\frac{d C}{d x}$$ = 0
⇒ – $$\frac{48}{x^2}$$ + 6x = 0
⇒ 6x3 – 48 = 0
⇒ x = 2
∴ ($$\frac{d^2 P}{d x^2}$$)x = 2 = $$\frac{96}{8}$$ + 6
= 12 + 6 = 18 > 0
Thus x = 2 be a point of minima.
and Minimum value of C = 5 + $$\frac{48}{2}$$ + 3 × 22
= 5 + 24 + 12 = Rs. 41

Question 10.
Prove that of all rectangles with given area, the square has the smallest perimeter.
Solution:
Let x and y be the length and breadth of rectangle.
∴ A = area of rectangle = xy [given] ………..(1)
Let P = perimeter = 2 (x + y)
= 2 [x + $$\frac{A}{x}$$] [using (1)]
∴ $$\frac{d P}{dx}$$ = 2 (1 – $$\frac{\mathrm{A}}{x^2}$$)
⇒ $$\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{x^3}$$
For maxima or minima, $$\frac{d P}{dx}$$ = 0
⇒ 1 – $$\frac{\mathrm{A}}{x^2}$$ = 0
⇒ x = √A [∵ x > 0]
∴ $$\frac{d^2 \mathrm{P}}{d x^2}=\frac{4 \mathrm{~A}}{\mathrm{~A}^{3 / 2}}=\frac{4}{\sqrt{\mathrm{A}}}$$ > 0
∴ P is minimise for x = √A
∴ From (1) ;
y = $$\frac{\mathrm{A}}{\sqrt{\mathrm{A}}}$$ = √A
∴ x = y
Hence the rectangle becomes square.
Thus, the rectangle with given area, the square has the smallest perimeter.

Question 11.
Of all rectangles each of which has perimeter 24 cm, find the one having maximum area. Also find that area.
Solution:
Let x cm be the length and y cm be the breadth of rectangle
s.t 2(x + y) = 24
[∵ perimeter of rectangle = 24]
⇒ x + y = 12
Let A = area of rectangle = xy = x (12 – x) [using (1)]
Differentiating w.r.t. x, we have
$$\frac{d A}{d x}$$ = 12 – 2x;
$$\frac{d^2 A}{d x^2}$$ = – 2
For maxima/minima, $$\frac{d A}{d x}$$ = 0
⇒ 12 – 2x = 0
⇒ x = 6
∴ ($$\frac{d^2 A}{d x^2}$$)x = 6 = – 2 < 0
∴ x = 6 be a point of maxima.
Thus Area A is maximum for x = 6
∴ from (1);
y = 12 – 6 = 6
Therefore x = y = 6 cm.
Thus the square has the maximum area.
and Maximum area = xy = (6 × 6) cm2 = 36 cm2.

Question 12.
Prove that of all rectangles with given perimeter, the square has the largest area.
Solution:
Let x cm and y cm be the length and breadth of the rectangle.
Let P be the perimeter of rectangle
∴ P = 2x + 2y …………(1)
Let A = area of rectangle = xy
= x ($$\frac{P}{2}$$ – x)
⇒ A = $$\frac{P x}{2}$$ – x2
Diff, both sides w.r.t. x; we have
$$\frac{d A}{d x}$$ = $$\frac{P}{2}$$ – 2x ;
$$\frac{d^2 A}{d x^2}$$ = – 2
For maxima/minima, $$\frac{d A}{d x}$$ = 0
⇒ $$\frac{P}{2}$$ – 2x = 0
⇒ x = $$\frac{P}{4}$$
∴ ($$\frac{d^2 A}{d x^2}$$)x = $$\frac{P}{4}$$ = – 2 < 0
Thus, x = $$\frac{P}{4}$$ be a point of maxima
∴ from (1) ;
2y = P – $$\frac{P}{2}$$ = $$\frac{P}{2}$$
⇒ y = $$\frac{P}{4}$$
Thus area of the rectangle is maximum when x = y = $$\frac{P}{4}$$
∴ rectangle becomes square.
Thus, Maximum area = $$\frac{P}{4} \times \frac{P}{4}=\frac{P^2}{16}$$
Hence, the rectangle with given perimeter, the square has the largest area.

Question 13.
Prove that the perimeter of a right- angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Solution:
Let ΔABC be right angled Δ at angle B with hypotenuse h.
Let AB = x
∴ BC = $$\sqrt{h^2-x^2}$$
[using pythagoras theorem]

Thus P = perimetcr of ∆ABC = AB + BC + AC
⇒ P = x + $$\sqrt{h^2-x^2}$$ + h
Diff. (1) both sides w.r.t. x; we have
$$\frac{d \mathrm{P}}{d x}=1+\frac{1}{2} \frac{1}{\sqrt{h^2-x^2}}(-2 x)$$
= 1 – $$\frac{x}{\sqrt{h^2-x^2}}$$
For maxima / minima, $$\frac{d P}{d x}$$ = 0
⇒ 1 – $$\frac{x}{\sqrt{h^2-x^2}}$$ = 0
⇒ $$\sqrt{h^2-x^2}$$ = x
⇒ h2 – x2 = x2
⇒ 2x2 = h2
⇒ x = ± $$\frac{h}{\sqrt{2}}$$
Since x, h > 0
∴ x = $$\frac{h}{\sqrt{2}}$$

∴ AB = BC
Hence the perimeter of right angle triangle is maximum when the triangle is isosceles.

Question 14.
Two sides of a triangle have lengths a and b, and the angle between them is 6. What value of 9 will maximize the area of the triangle ? Also find the maximum area of the triangle.
Solution:
Let ABC be the given triangle with AC = b
and BC = a be given two sides.
Let θ be the angle between them
Let A = area of ∆ABC
= $$\frac{1}{2}$$ ab sin θ

Clearly A is maximum when sin θ is maximum i.e. sin θ = 1
⇒ θ = $$\frac{\pi}{2}$$
and required maximum area = $$\frac{1}{2}$$ ab × 1 = $$\frac{1}{2}$$ ab

Question 15.
A right-angled triangle with constant area is given. Prove that the hypotenuse of the triangle is least when the triangle is-isosceles.
Solution:
Let ∆ABC be right angled triangle with constant area P.
Let l be the length of the hypotenuse of ∆ABC
∴ AB = l sin θ
and BC = l cos θ
Also ∠ACB = θ.
∴ area of triangle ∆ABC = $$\frac{1}{2}$$ × BC × AB
⇒ P = $$\frac{1}{2}$$ (l cos θ) (l sin θ) = $$\frac{l^2}{4}$$ sin 2θ
⇒ l2 = 4P cosec 2θ
Now l is minimise when l2 is minimise, i.e. we want to find the value of 0 for which l2 is least.
Let L = l2 = 4P cosec 2θ
Diff. both sides w.r.t. θ, we have
$$\frac{d \mathrm{~L}}{d \theta}$$ = – 8P cosec 2θ cot 2θ
and $$\frac{d^2 \mathrm{~L}}{d \theta^2}$$ = – 8P [- 2 cosec3 2θ + cot 2θ (- 2 cosec 2θ cot 2θ)]
= 16P [cosec3 2θ + cosec 2θ cot2 2θ]
For maxima/minima $$\frac{d \mathrm{~L}}{d \theta}$$ = 0
⇒ cosec 2θ cot 2θ = 0
⇒ $$\frac{\cos 2 \theta}{\sin ^2 2 \theta}$$ = 0
⇒ cos 2θ = 0
⇒ 2θ = $$\frac{\pi}{2}$$
⇒ θ = $$\frac{\pi}{4}$$
∵ θ ∈ (0, $$\frac{\pi}{2}$$)
At θ = $$\frac{\pi}{4}$$,
$$\frac{d^2 \mathrm{~L}}{d \theta^2}$$ = 16P [13 + 1 × 0] = 16P > 0
∴ Lis minimum for θ = $$\frac{\pi}{4}$$
Thus l is least for θ = $$\frac{\pi}{4}$$
∴ AB = l sin $$\frac{\pi}{4}$$
= $$\frac{l}{\sqrt{2}}$$
and BC = l cos θ
= l cos $$\frac{\pi}{4}$$
= $$\frac{l}{\sqrt{2}}$$
∴ AB = BC.
Hence ∆ABC be an isosceles triangle.

Question 16.
The lengths of the sides of a triangle are 9 + x2, 9 + x2 and 18 – 2x2. Calculate:
(i) the area of the triangle in terms of x.
(ii) the value of x for which this area is maximum.
Solution:
(i) Here given lengths of sides of triangle are 9 + x2, 9 + x2 and 18 – 2x2.
Let a = 9 + x2 ;
b = 9 + x2;
c = 18 – 2x2
∴ s = $$\frac{a+b+c}{2}$$
= $$\frac{9+x^2+9+x^2+18-2 x^2}{2}$$ = 18
Let A = area of ∆ABC
= $$\sqrt{s(s-a)(s-b)(s-c)}$$ [By Heron’s formula]
= $$\sqrt{18\left(18-9-x^2\right)\left(18-9-x^2\right)\left(18-18+2 x^2\right)}$$
= $$\sqrt{18\left(9-x^2\right)^2 \times 2 x^2}$$
= 6 (9 – x2) x
= 54x – 6x3

(ii) Diff, both sides w.r.t. x, we have
$$\frac{d A}{d x}$$ = 6 [9 – 3x2] ;
$$\frac{d^2 A}{d x^2}$$ = 6 (0 – 6x) = – 36x
For maxima / minima, $$\frac{d A}{d x}$$ = 0
⇒ $$\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{3}}$$ = – 36√3 < 0
Thus, x = √3 be a point of maxima
and maximum value of area A = 54 × √3 – (6√3)3
= 54√3 – 18√3 = 36√3 sq. units.

Question 17.
The perimeter of a triangle is 8 cm. If one of the sides is 3 cm, what are the lengths of the other sides for maximum area of the triangle ?
Solution:
Let a, b, c be the lengths of sides of triangle then a + b + c = 8
∴ s = $$\frac{a+b+c}{2}=\frac{8}{2}$$ = 4
given one of its sides be of length 3 cm,
let a = 3
b + c = 5
⇒ b = 5 – c
Let A = area of the ∆ABC
= $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{4(4-3)(4-5+c)(4-c)}$$
= $$\sqrt{4(c-1)(4-c)}$$
⇒ A2 = 4 (c – 1) (4 – c) = 4 (- c2 + 5c – 4)
Now A is maximise when A2 is maximise.
Let P = A2 = 4 (- c2 + 5c – 4)
On differentiating w.r.t. c, we get
$$\frac{d P}{d c}$$ = 4 (- 2c + 5) ;
∴ ($$\frac{d^2 P}{d c^2}$$)c = $$\frac{5}{2}$$ = – 8 < 0
Thus P is maximum for c = $$\frac{5}{2}$$
Hence Area A is maximum for c = $$\frac{5}{2}$$
Thus the required lengths of sides of triangle are 3, 5 – $$\frac{5}{2}$$ and $$\frac{5}{2}$$ i.e. 3 cm, $$\frac{5}{2}$$ cm and $$\frac{5}{2}$$ cm.

Question 18.
A sheet of paper is to contain 18 cm2 of printed matter. The margins at the top and bottom are 2 cm each, and at the sides 1 cm each. Find the dimensions of the sheet which require the least amount of paper.
Solution:
Let x cm (x > 0) be the one dimension of the page then the other dimension be $$\frac{18}{x}$$ cm, since the area of given page be 18 sq. cm.

Let A cm2 be the area of printed page.
Then, A = (x – 2) ($$\frac{18}{x}$$ – 4)
= 18 – 4x – $$\frac{36}{x}$$ + 8
= 26 – 4x – $$\frac{36}{x}$$
On differentiating w.r.t. x, we have
$$\frac{d \mathrm{~A}}{d x}=-4+\frac{36}{x^2}$$
$$\frac{d^2 \mathrm{~A}}{d x^2}=-\frac{72}{x^3}$$
For maxima / minima, $$\frac{d A}{d x}$$ = 0
⇒ – 4 + $$\frac{36}{x^2}$$ = 0
⇒ x2 = 9
⇒ x = 3
∴ ($$\frac{d^2 A}{d x^2}$$)x = 3 = $$-\frac{72}{27}=-\frac{8}{3}$$ < 0
Thus A is maximum for x = 3
∴ the dimensions of the printed page be x and $$\frac{18}{x}$$ i.e. 3 cm and 6 cm
Hence, the required dimensions of the sheet that requires the least amount of paper be (x + 2) and ($$\frac{18}{x}$$ + 4) cm i.e. 5 cm and 10 cm.

Question 19.
Show that of all rectnagles inscribed in a given fixed circle the square has the maximum area.
Solution:
Let x and y be the length and breadth of the rectangle respectively.
Which is inscribed in a circle of radius r
∴ x2 + y2 = (2r)2
x2 + y2 = 4r2 …………(1)
Thus A = area of rectangle = xy
= x $$\sqrt{4 r^2-x^2}$$ [using (1)]

For max/ minima, $$\frac{d A}{d x}$$ = 0
⇒ 4r2 – 2x2 = 0
⇒ x = √2r
[∵ x > 0]
∴ $$\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}$$ < 0
∴ A is maximum for x = √2r
∴ From (1) ;
2r2 + y2 = 4r2
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20.
(i) The sum of perimeters of a circle and a square is k, where k is some constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy
Let A combined area of square and circle
⇒ A = x2 + πy2 [using (1)]
= x2 + π $$\left(\frac{\mathrm{P}-4 x}{2 \pi}\right)^2$$ [using (1)]
⇒ A = x2 + $$\frac{1}{4 \pi}$$ (P – 4x)2
∴ $$\frac{d A}{d x}$$ = 2x + $$\frac{1}{2 \pi}$$ (P – 4x) (- 4)
For maxima / minima, we have $$\frac{d A}{d x}$$ = 0
⇒ 4r2 – 2x2 = 0
⇒ x = √2r [∵ x > 0]
∴ $$\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\sqrt{2} r}=\frac{-\sqrt{2} r\left(8 r^2\right)}{\left(2 r^2\right)^{3 / 2}}$$ < 0
∴ A is minimum for x = √2r
∴ From (1) ;
2r2 + y = 4r2
⇒ y = √2r
Hence A is maximise for x = y = √2r
i.e. rectangle becomes square.

Question 20 (old).
(i) The sum of perimeters of a circle and a square is k, where A is some constant. Prove that the sum of their arcas is least when the side of the square is double the radius of the circle. (NCERT)
(ii) Given the sum of perimeters of a circle and a square, show that the sum of areas is least when the diameter of the circle is equal to the side of the square.
Solution:
(i) Let x be the side of square and y be the radius of circle.
Let P be the sum of their perimeters.
Then P = 4x + 2πy …………..(1)
Let A = combined area of square and circle
⇒ A = x2 + πy2
= x2 + π $$\left(\frac{P-4 x}{2 \pi}\right)^2$$ [using (1)]
⇒ A = x2 + $$\frac{1}{4 \pi}$$ (P – 4x)2
∴ $$\frac{d A}{d x}$$ = 2x + $$\frac{1}{2 \pi}$$ (P – 4x) (- 4)
For maxima/minima, we have $$\frac{d A}{d x}$$ = 0
2x – $$\frac{2}{\pi}$$ (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ x = $$\frac{2 P}{2 \pi+8}=\frac{P}{\pi+4}$$
Also $$\frac{d^2 P}{d x^2}$$ = 2 + $$\frac{16}{2 \pi}$$ > 0
∴ P is minimum for x = $$\frac{9}{\pi+4}$$
∴ from (1) ;
2πy = P – $$\frac{4 \mathrm{P}}{\pi+4}$$
⇒ 2πy = $$\frac{\pi \mathrm{P}}{\pi+4}$$
⇒ y = $$\frac{\mathrm{P}}{2(\pi+4)}=\frac{x}{2}$$
⇒ x = 2y
i.e. side of square = diameter of circle.
Hence A is minimum when side of the square is equal to the diameter of the circle.

(ii) Let x be the length of each side of square and r be the radius of circle.
Then P = sum of perimeters of a circle and as square = 2πr + 4x …………(1)
Let A = combined area of square and circle
⇒ A = x2 + r2
= x2 + π $$\left(\frac{P-4 x}{2 \pi}\right)^2$$ [using (1)]
⇒ A = x2 + $$\frac{1}{4 \pi}$$ (P – 4x)2 ;
Diff. both sides w.r.t. x, we have
$$\frac{d A}{d x}$$ = 2x + $$\frac{1}{4 \pi}$$ × 2 (P – 4x) (- 4)
and $$\frac{d^2 A}{d x^2}$$ = 2 + (- $$\frac{2}{\pi}$$) (- 4)
= 2 + $$\frac{8}{\pi}$$
For maxima and minima, $$\frac{d A}{d x}$$ = 0
⇒ 2x – $$\frac{2}{\pi}$$ (P – 4x) = 0
⇒ 2πx – 2P + 8x = 0
⇒ (π + 4) x = P
⇒ x = $$\frac{P}{\pi+4}$$
Now, $$\left(\frac{d^2 \mathrm{~A}}{d x^2}\right)_{x=\frac{\mathrm{P}}{\pi+4}}$$ = 2 + $$\frac{8}{\pi}$$
Hence A is least when x = $$\frac{P}{\pi+4}$$
⇒ (π + 4) x = 2πr + 4x [using (1)]
⇒ πx = 2πr
⇒ x = 2r
Thus, diameter of circle is equal to side of square.

Question 21.
A wire 10 metres long is cut into two parts. One part is bent into the shape of a circle and the other into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the two figures is as small as possible?
Solution:
Let r metre be the radius of circle and a metres be the side of equilateral triangle.
Then 2πr + 3a = 10 ………….( 1)
Let A combined area of circle and equilateral triangle
⇒ A = πr2 + $$\frac{\sqrt{3}}{4}$$ a2
= πr2 + $$\frac{\sqrt{3}}{4}\left(\frac{10-2 \pi r}{3}\right)^2$$ [using (1)]
On differentiating both sides w.r.t. r, we have
$$\frac{d A}{d r}$$ = 2πr + $$\frac{\sqrt{3}}{36}$$ 2 (10 – 2πr) (- 2π)
= 2πr – $$\frac{\sqrt{3}}{9}$$ π (10 – 2πr)
∴ $$\frac{d^2 A}{d x^2}$$ = 2π – $$\frac{\sqrt{3} \pi}{9}$$ (0 – 2π)
= 2π + $$\frac{2 \sqrt{3}}{9}$$ π2
For maxima / minima, $$\frac{d A}{d r}$$ = 0
⇒ 2πr – $$\frac{\sqrt{3} \pi}{9}$$ (10 – 2πr) = 0
⇒ 18πr – 10√3π + 2√3π2r = 0
⇒ (18 + 2√3π) r = 10√3
⇒ r = $$\frac{10 \sqrt{3}}{18+2 \sqrt{3} \pi}$$
⇒ r = $$\frac{10 \sqrt{3}}{2 \sqrt{3}(\pi+3 \sqrt{3})}$$
= $$\frac{5}{\pi+2 \sqrt{3}}$$ m
Also, at r = $$\frac{5}{\pi+3 \sqrt{3}}$$ ;
$$\frac{d^2 A}{d x^2}$$ = 2π + $$\frac{2 \sqrt{3}}{9}$$ π2 > 0
∴ A is least for r = $$\frac{5}{\pi+3 \sqrt{3}}$$ m
Thus, length of piece of wire bent into the form of circle = 2πr = $$\frac{10 \pi}{\pi+3 \sqrt{3}}$$ m
and length of piece of wire bent into form of equilateral triangle = $$\left(10-\frac{10 \pi}{\pi+3 \sqrt{3}}\right)$$ m
= $$\frac{30 \sqrt{3}}{\pi+3 \sqrt{3}}$$ m

Question 22.
A wire of length 36 cm is cut into two pieces. One of the piece is turned into the form of a square and the other in the form of an equilateral triangle. Find the lengh of each piece so that the sum of the areas of the two figures be minimum.
Solution:
Let the length of one piece of wire be x m
∴ other piece must be (36 – x) m.
Let x metres be made into square and (36 – x) m be made into equilateral ∆.
So perimeter of square = 4 × side = x
∴ side of square = $$\frac{x}{4}$$
∴ area of square = $$\left(\frac{x}{4}\right)^2$$
Further perimeter of equilateral ∆ = 3 × side
= 36 – x
∴ side of equilateral ∆ = 12 – $$\frac{x}{3}$$
∴ area of equilateral ∆ = $$\frac{\sqrt{3}}{4}$$ (side)2
= $$\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2$$
∴ A = combined area of square and equilateral ∆ = $$\frac{x^2}{16}+\frac{\sqrt{3}}{4}\left(12-\frac{x}{3}\right)^2$$
∴ $$\frac{d A}{d x}$$ = $$\frac{x}{8}+\frac{\sqrt{3}}{2}\left(12-\frac{x}{3}\right)\left(-\frac{1}{3}\right)$$
$$\frac{d A}{d x}$$ = $$\frac{x}{8}+\frac{x}{6 \sqrt{3}}$$ – 2
For maxima / minima, $$\frac{d A}{d x}$$ = 0

Hence the length of two pieces are $$\frac{144}{4+3 \sqrt{3}}$$ m and $$\left(36-\frac{144}{x+3 \sqrt{3}}\right)$$ m i.e. $$\frac{108 \sqrt{3}}{4+3 \sqrt{3}}$$ m.

Question 22.
A window is in the form of a rectangle surrounded by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to adniit maximum light through the whole opening.
Solution:
Let the width and height of the window be 2x metre and y metre respectively.
Given total perimeter of window = 10 m
⇒ 2x + 2y + πx = 10 ………..(1)
Let A = total area of window
⇒ A = 2xy + $$\frac{\pi x^2}{2}$$
⇒ A = x (10 – πx – 2x) + $$\frac{\pi x^2}{2}$$
∴ $$\frac{d A}{d x}$$ = 10 – (π + 2) 2x + πx

For maxima/minima, $$\frac{d A}{d x}$$ = 0
⇒ 10 – 2πx – 4x + πx = 0
⇒ 10 – 4x – πx = 0
⇒ x = $$\frac{10}{\pi+4}$$
∴ $$\frac{d^2 A}{d x^2}$$ = – 2 (π + 2)
= – π – 4
at x = $$\frac{10}{\pi+4}$$;
$$\frac{d^2 A}{d x^2}$$ = – π – 4 < 0
Hence A is maximum when x = $$\frac{10}{\pi+4}$$
∴ from (1) ;
2y = 10 – $$\frac{10}{\pi+4}$$
= $$\frac{10 \pi+40-10 \pi-20}{\pi+4}$$
⇒ y = $$\frac{10}{\pi+4}$$
Thus, dimensions of the window are 2x and y
i.e. $$\frac{2 \times 10}{\pi+4}$$ and $$\frac{10}{\pi+4}$$
i.e. $$\frac{10}{\pi+4}$$ and $$\frac{10}{\pi+4}$$.

Question 23.
A window is in the form of a rectangle above which there is a semicircle. If the perimeter of the window is p cm, show that the window will admit maximum possible light only when the radius of semicircle is $$\frac{p}{\pi+4}$$ cm.
Solution:
Let r cm be the radius of semi-circle and x cm be the side BC of rectangle as shown in figure shown alongside.
Then p perimeter of combined figure
⇒ p = πr + 2x + 2r ……….(1)

Let A = area of the window.
⇒ A = $$\frac{1}{2}$$ πr2 + 2rx
= $$\frac{\pi r^2}{2}+2 r \frac{(p-\pi r-2 r)}{2}$$ ………(2) [using (1)]
Now maximum light wil be admitted through the window if the area of the window is maximum.
For this we have to maximise A.
Diff. eqn. (2) both sides w.r.t. r, we have
$$\frac{d A}{d r}$$ = πr
For maxima/minima, $$\frac{d A}{d r}$$ = 0
⇒ πr + p – 2πr – 4r = 0
⇒ p – πr – 4r = 0
⇒ r = $$\frac{p}{\pi+4}$$
∴ at r = $$\frac{p}{\pi+4}$$,
$$\frac{d^2 A}{d x^2}$$ = – (π + 4) < 0
Thus A is least for r = $$\frac{p}{\pi+4}$$
Thus, from eqn. (1); we have
2x = p – $$\frac{(\pi+2) p}{\pi+4}$$
= $$\frac{p \pi+4 p-\pi p-2 p}{\pi+4}$$
⇒ x = $$\frac{p}{\pi+4}$$
Hence, maximum light is admitted when radius of semi-circle be r = $$\frac{p}{\pi+4}$$ cm.

Question 24.
A rectangular area of 9000 m2 is to be surrounded by a fence with two opposite sides being made of brick and the other two of wood. One metre of wooden fencing costs 25 while one metre of brick walling costs 10. What is the least amount of money that must be alloted for the construction of such a fence?
Solution:
Let x be the length of rectangular area and
y be the breadth of rectangula area
∴ xy = 9000 m2
Let C = amount of money that must be allotted for the construction of such a fence.
∴ C = 25 × 2x + 2y × 10
= 50x + 20 × $$\frac{9000}{x}$$ [using (1)]
Diff. both sides w.r.t. x; we have
$$\frac{d C}{d x}$$ = 50 – $$\frac{180000}{x^2}$$ ;
$$\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}$$
For maxima / minima, $$\frac{d C}{d x}$$ = 0
⇒ 50 – $$\frac{180000}{x^2}$$ = 0
⇒ x2 = $$\frac{18000}{50}$$ = 3600
⇒ x = 60 (∵ x > 0)
∴ from (1) ;
⇒ y = $$\frac{9000}{60}$$ = 150
Now $$\frac{d^2 \mathrm{C}}{d x^2}=\frac{360000}{x^3}$$
∴ $$\left(\frac{d^2 \mathrm{C}}{d x^2}\right)_{x=60}=\frac{360000}{60 \times 60 \times 60}$$
= $$\frac{360}{216}=\frac{5}{3}$$ > 0
Thus C is minimise for x = 60 and y = 150
∴ required amount of money = C
= ₹ (50 × 60 + 20 × 150)
= ₹ 6000.

Question 25.
(i) Find the point on the parabola y2 = 4x which is nearest to the point (2, – 8).
(ii) Find the point on the curve x2 = 8y which is nearest to the point (2, 4).
Sol.
(I) Let P(x, y) be any point on the curve
y2 = 4x …………(1)
Let A (2, – 8) be the given point.
∴ |AP| = $$\sqrt{(x-2)^2+(y+8)^2}$$
⇒ AP2 = (x – 2)2 + (y + 8)2
= ($$\frac{y^2}{4}$$ – 4)2 + (y + 8)2 [using (1)]
Let S = AP2,
Then S is maximum or minimum according as AP is maximum or minimum.
Now, S = ($$\frac{y^2}{4}$$ – 4)2 + + (y + 8)2
∴ $$\frac{d S}{d y}$$ = 2 ($$\frac{y^2}{4}$$ – 2) ($$\frac{y}{2}$$) + 2 (y + 8)
∴ $$\frac{d S}{d y}$$ = $$\frac{y^3}{4}$$ – 2y + 2y + 16
= $$\frac{y^3}{4}$$ + 16
∴ $$\frac{d^2 \mathrm{~S}}{d y^2}=\frac{3 y^2}{4}$$
For max./minima, $$\frac{d S}{d y}$$ = 0
⇒ y3 = – 64
⇒ y3 = – 64
⇒ y = – 4
∴ ($$\frac{d^2 S}{d y^2}$$)y = – 4 = $$\frac{3}{4}$$ (16) = 12 > 0
∴ S is minimise for y = – 4
and from (1); x = 4
Hence the point (4, – 4) on curve y32 = 4x is nearest to the given point (2, – 8).

(ii) Let P(x, y) be any point on the curve
x2 = 8y …………………..(1)
and let the given point be A (2, 4).
∴ AP2 = (x – 2)2 + (y – 4)2
= (x – 2)2 + $$\left(\frac{x^2}{8}-4\right)^2$$ [using (1)]
Let S = AP2,
Now S is max./min.
according as AP is max./min.
∴ S = (x – 2)2 + $$\left(\frac{x^2}{8}-4\right)^2$$
∴ $$\frac{d S}{d x}$$ = 2 (x – 2) + 2 $$\left(\frac{x^2}{8}-4\right) \frac{2 x}{8}$$
= 2x – 4 + $$\frac{x^3}{16}$$ – 2x
∴ $$\frac{d \mathrm{~S}}{d x}=\frac{x^3}{16}$$ – 4
∴ $$\frac{d^2 \mathrm{~S}}{d x^2}=\frac{3 x^2}{16}$$
For max./min., $$\frac{d S}{d x}$$ = 0
⇒ x3 = 8
⇒ x = 4
when x = 4,
$$\left(\frac{d^2 \mathrm{~S}}{d x^2}\right)=\frac{3 \times 16}{16}$$ = 3 > 0
∴ S is minimize for x = 4
∴ from (1) ; y = 2
∴ Required point on given curve be (4, 2).

Question 26.
Find the maximum area of an isosceles triangle inscribed in the ellipse $$\frac{x^2}{16}+\frac{y^2}{9}$$ = 1 with its vertex at one end of the major axis.
Solution:
Let any point P on the ellipse

$$\frac{x^2}{16}+\frac{y^2}{9}$$ = 1 is (4 cos θ, 3 sin θ)
∴ A = Area of ∆ $$\frac{1}{2}$$ (PP’) (AB)
= $$\frac{1}{2}$$ (6 sin θ) (4 – 4 cos θ)
= 12 sin θ (1 – cos θ)
= 12 [sin θ – $$\frac{1}{2}$$ sin 2θ]
∴ $$\frac{d \mathrm{~A}}{d \theta}$$ = 12 [cos θ – cos 2θ]
For Maxima / Minima, $$\frac{d \mathrm{~A}}{d \theta}$$ = 0
⇒ cos 2θ – cos θ = 0
⇒ 2 cos θ – cos θ – 1 = 0
⇒ (cos θ – 1) (2 cos θ + 1) = 0
⇒ cos θ = 1 or
cos θ = – $$\frac{1}{2}$$ = cos (π – $$\frac{\pi}{3}$$)
⇒ θ = 0 or θ = $$\frac{2 \pi}{3}$$
When θ = 0, the point P coincide with A (4, 0)
When θ = $$\frac{2 \pi}{3}$$,
Now $$\frac{d^2 \mathrm{~A}}{d \theta^2}$$ = 12 (- sin θ + 2 sin 2θ)
⇒ $$\left(\frac{d^2 \mathrm{~A}}{d \theta^2}\right)_{\theta=\frac{2 \pi}{3}}=12\left(-\sin \frac{2 \pi}{3}+2 \sin \frac{4 \pi}{3}\right)$$
= $$\left(-\frac{\sqrt{3}}{2}+2\left(\frac{-\sqrt{3}}{2}\right)\right)$$
= – 18√3 < 0
∴ A is maximise at θ = $$\frac{\pi}{3}$$
Hence maximum area = A
= 12 $$\left[\sin \frac{2 \pi}{3}-\frac{1}{2} \sin \frac{4 \pi}{3}\right]$$
= 12 $$\left[\frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{\sqrt{3}}{2}\right]$$
= 9√3 sq. units

Question 27.
Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.
Solution:
Let x, x, y are the length, breadth and height of closed cuboid.
∴ area of base square = x2 ;
area of four walls = 4xy
Let S = area of cuboid = 2x2 + 4xy ……….(1)
Further, V = volume of cuboid = x2y ………..(2)
∴ S = 2x2 + 4x $$\left(\frac{\mathrm{V}}{x^2}\right)$$ [using (1)]
∴ $$\frac{d \mathrm{~S}}{d x}=4 x-\frac{4 \mathrm{~V}}{x^2}$$
∴ $$\frac{d^2 \mathrm{~S}}{d x^2}=4+\frac{8 \mathrm{~V}}{x^3}$$
For Max/Min, $$\frac{d S}{d x}$$ = 0
⇒ 4x = $$\frac{4 V}{x^2}$$
⇒ x3 = V
⇒ x = (V)1/3
at x = (V)1/3
⇒ $$\frac{d^2 S}{d x^2}$$ = 4 + $$\frac{8 V}{V}$$ = 12 > 0
∴ S is minimise for x = $$\sqrt[3]{v}$$
∴ y = $$\frac{\mathrm{V}}{x^2}=\frac{\mathrm{V}}{\mathrm{V}^{2 / 3}}$$
= V1/3
Hence all the dimensions of cuboid are equal.
∴ Cuboid because cube.
Hence the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

Question 28.
An open box with a square base is to be made of given iron sheet of area 27 sq.m. Show that the maximum volume of the box is 13.5 cu. m .
Solution:
Let x, x and y are the length, breadth and height of square box.
It is given that
x2 + 4xy = 27 sq. m
and V = volume of box = x2y
⇒ V = x2 $$\left(\frac{27-x^2}{4 x}\right)$$
= $$\frac{27}{4} x-\frac{x^3}{4}$$
∴ $$\frac{d V}{d x}=\frac{27}{4}-\frac{3 x^2}{4}$$
For maximaJminima,
⇒ $$\frac{d V}{d x}$$ = 0
⇒ $$\frac{27}{4}-\frac{3 x^2}{4}$$ = 0
⇒ 27 = 3x2
⇒ at x = 3; (∵ x > 0)
Thus V is maximise volume x = 3
∴ y = $$\frac{27-9}{4 \times 3}=\frac{3}{2}$$
∴ Maximum volume of box = x2y
=32 × $$\frac{3}{2}$$
= $$\frac{27}{2}$$ cu.m = 13.5 cu.m.

Question 28 (old).
An open box with a square base is to be made of given ¡ron sheet of area 27 sq. m. Show that the maximum volume of the box is 13.5 cu. m.
Solution:
Let x be the side of the square base and y be the height of the box.
Thus area of square base = x2
and area of four walls = 4 xy
given area of square box + area of four val1s = 27 m2
⇒ x2 + 4xy = 27
Let V = volume of box = x2y = x2 $$\left(\frac{27-x^2}{4 x}\right)$$ [using (1)]
V = $$\frac{1}{4}$$ (27x – x3)
Diff, both sides w.r,t. x, we have
$$\frac{d V}{d x}$$ = $$\frac{1}{4}$$ (27 – 3x2) ;
$$\frac{d^2 V}{d x^2}$$ = $$\frac{1}{4}$$ (- 6x)
For maxima / minima, $$\frac{d V}{d x}$$ = 0
⇒ 27 – 3x2 = 0
⇒ x2 = 9
⇒ x = 3 (∵ x > 0)
∴ $$\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=3}=-\frac{3}{2} \times 3=-\frac{9}{2}$$ < 0
Thus V is minimise for x = 3
∴ from (1) ; we have
y = $$\frac{27-x^2}{4 x}$$
= $$\frac{27-9}{12}$$
= $$\frac{18}{12}$$
= $$\frac{3}{2}$$
Hence, the required maximum volume of box = x2y
= (32 × $$\frac{3}{2}$$) cu.m = 13.5 cu.m.

Question 29.
The volume of a closed metal box with a square base is 4096 cm3. The cost of polishing the outer surface of the box is Rs. 4 per cm2. Find the dimensions of the box for the minimum cost of polishing it. (ISC 2019)
Solution:
Let the dimensions of closed metal box are x cm, x cm and y cm respectively.
Then volume of closed metal box = x2y
and given volume of box = 4096 cm3
∴ x2y = 4096
Let S = surface area of closed metal box = 2x2 + 4xy.
Given, the cost of polishing the outer surface of box be Rs. 4 per cm2.
Let C be the cost of polishing the whole metal box
Then C = 4(2x2 + 4xy) = 8(x2 + 2xy)
C = 8(x2 + 2x × $$\frac{4096}{x^2}$$) [using eqn. (1)]
∴ $$\frac{d \mathrm{C}}{d x}=8\left(2 x-\frac{2 \times 4096}{x^2}\right)$$
= 16 $$\left(x-\frac{4096}{x^2}\right)$$
For maximum / minima, $$\frac{d C}{d x}$$ = 0
⇒ x3 = 4096
⇒ x = 16 (∵ x > 0)
Now $$\frac{d^2 \mathrm{C}}{d x^2}=16\left(1+\frac{2 \times 4096}{x^3}\right)$$
∴ $$\left(\frac{d^2 c}{d x^2}\right)_{x=16}=16\left(1+\frac{2 \times 4096}{16^3}\right)$$
= 48 > 0
Thus x = 16 be a point of minima
4096
∴ from (1) ;
y = $$\frac{4096}{16^2}$$ = 16
Thus, cost is minimum when x = 16 cm and y = 16 cm.

Question 30.
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width.
Solution:
Let x, x and y are the length, width and height of an open tank
∴ Volume of tank V = x2y (given) …………..(1)
Now it is given that, the tank to hold a given quantity of water
∴ V = constant.
To minimise the cost of material. we have to minimise the surface area of the tank.
Let S = x2 + 4xy
= x2 + 4x ($$\frac{\mathrm{~V}}{x^2}$$) (using (1))
⇒ $$\frac{d \mathrm{~S}}{d x}=2 x-\frac{4 \mathrm{~V}}{x^2}$$
∴ $$\frac{d^2 \mathrm{~S}}{d x^2}=2+\frac{8 \mathrm{~V}}{x^3}$$
For max./minima, $$\frac{d S}{d x}$$ = 0
⇒ x = (2V)1/3
and $$\frac{d^2 S}{d x^2}$$ = 2 + $$\frac{8 V}{2 V}$$ = 6 > 0
∴ S is minimise for x3 = 2V.
i.e. x3 = 2x2y
⇒ x = 2y
⇒ y = $$\frac{x}{2}$$
i.e. S is minimum when depth of the tank is half of its width.

Question 31.
A tank with rectangular sides, open at the top, is to be constructed so that its depth is 2 m and volume is 8 m3. 1f building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for the sides, what is the cost of least expensive tank ?
Solution:
Let x, x, y be the length, width and height of the tank
∴ area of square base = x2 and
area of four walls = 4xy
Given V = volume of open tank = 8 x2y …………(1)
∴ E = 70x2 + 45 (4xy)
= 70x2 + 180x . $$\frac{8}{x^2}$$ [using (1)]
⇒ E = 70x2 + $$\frac{180 \times 8}{x}$$ ……….(2)
Given y = depth of tank = 2m
∴ from (1) ;
8 = x2 × 2
⇒ x = 2[∵ x > 0]
Now we check whether x = 2 is point of maximum or minima.
$$\frac{d E}{d x}$$ = 140x – $$\frac{180 \times 8}{x^2}$$
$$\frac{d^2 \mathrm{E}}{d x^2}=140+\frac{180 \times 16}{x^2}$$
at x = 2,
$$\frac{d^2 E}{d x^2}$$ > 0
∴ E is minimum for x = 2.
∴ from (2) ; we have
least value of E = 70 (2)2 + $$\frac{180 \times 8}{2}$$ = ₹ 1000.

Question 32.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top by cutting off squares from the corners and folding up the flaps. What should be the side of the in square in order that the volume of the box is maximum ? (NCERT)
Solution:
Let x be the side of the square that is cut off from each comer of the plate.
Then sides of the base are 45 – 2x, 24 – 2x, x cm.
∴ Volume of the box = V = (45 – 2x) (24 – 2x) x
∴ V = (45 – 2x) (24x – 2x2)
= 2x (12 – x) (45 – 2x)
∴ V = 2x [540 – 24x – 45x + 2x2]
= 2x [2x2 – 69x + 540]
∴ $$\frac{d V}{d x}$$ = 2 [6x2 – 138x + 540]
= 4 [3x2 – 69x + 270]

For max/minima, $$\frac{d V}{d x}$$ = 0
3x2 – 69x + 270 = 0
∴ x = $$\frac{69 \pm 39}{6}$$ = 5, 18
But x ≠ 18.
Thus, x = 5,
$$\frac{d^2 V}{d x^2}$$ = 4 (6x – 69)
∴ ($$\frac{d^2 V}{d x^2}$$)x = 5 = 4 (- 39)
= – 156 < 0
∴ V is maximise for x = 5.
Hence the volume of the box is maximum when the side of square is 5 cm and max. volume = 35 × 14 × 5 = 2450 cm3.

Question 33.
Show that the semi-vertical angle of a cone of the maximum volume and of given slant height is cos-1 $$\frac{1}{\sqrt{3}}$$.
Solution:
Let θ be the semi-vertical angle of cone and / be the given slant height of cone.
∴ r = radius of cone = l sin θ
and h = height of cone = l cos θ
Let V = volume of cone = $$\frac{1}{3}$$ πr2h
= $$\frac{1}{3}$$ π (l sin θ)2 (l cos θ)

⇒ $$\frac{d V}{d \theta}$$ = $$\frac{\pi l^3}{3}$$ [sin2 θ (- sin θ) + 2 cos2 θ sin θ]
= $$\frac{\pi l^3}{3}$$ sin θ [2 cos2 θ – sin2 θ]
∴ $$\frac{d^2 \mathrm{~V}}{d \theta^2}$$ = $$\frac{\pi l^3}{3}$$ [- 3 sin2 θ cos θ + 2 cos3 θ – 4 cos θ sin2 θ]
For Max/minima, $$\frac{d V}{d \theta}$$ = 0
⇒ sin θ (2 cos2 θ – sin2 θ) = 0
∴ sin θ = 0
⇒ θ = 0, π which is not possible, (∵ in this case, r = 0)
∴ tan2 θ = 2
⇒ tan θ = √2 [∵ θ lies between 0 and π/2]
∴ θ = tan-1 (√2)
at tan θ = √2,

sin θ = $$\frac{\sqrt{2}}{\sqrt{3}}$$
and cos θ = $$\frac{1}{\sqrt{3}}$$
$$\frac{d^2 \mathrm{~V}}{d \theta^2}=\frac{\pi l^3}{3}\left[2 \times \frac{1}{3 \sqrt{3}}-7 \times \frac{1}{\sqrt{3}} \times \frac{2}{3}\right]$$
= $$\frac{\pi l^3}{3} \times\left(-\frac{12}{3 \sqrt{3}}\right)$$ < 0
∴ V is maximise for θ = tan-1 (√2) i.e. cos-1 $$\frac{1}{\sqrt{3}}$$

Question 34.
Show that a right circular cylinder of given volume, which is open at the top, has minimum total surface area if its height is equal to radius of its base.
Solution:
Let h be the height and r be the radius of cylinder
V = πr2h (given) ………….(1)
Let S = Total surface area
= πr2 + 2πrh
= πr2 + 2πr $$\frac{\mathrm{V}}{\pi r^2}$$ [using (1)]
∴ S = πr2 + $$\frac{2 V}{r}$$
∴ $$\frac{d S}{d r}$$ = 2πr – $$\frac{2 V}{r^{2}}$$
For Maxima / Minima, $$\frac{d S}{d r}$$ = 0
⇒ 2πr = $$\frac{2 V}{r^{2}}$$
⇒ r = $$\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}$$
Now $$\frac{d^2 S}{d r^2}$$ = 2π + $$\frac{4 V}{r^{3}}$$
∴ at r = $$\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}$$ ;
$$\frac{d^2 S}{d r^2}$$ = 2π + $$\frac{4 \mathrm{~V} \times \pi}{\mathrm{V}}$$ = 6π > 0
∴ S is minimise for r = $$\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}$$
∴ From (1) ; we have
h = $$\frac{\mathrm{V}}{\pi\left(\frac{\mathrm{V}}{\pi}\right)^{2 / 3}}$$
= $$\frac{\mathrm{V}^{1 / 3}}{\pi^{1 / 3}}=\left(\frac{\mathrm{V}}{\pi}\right)^{1 / 3}$$
Hence S is minimise when radius of cylinder is equal to height of cylinder.

Question 34 (old).
Show that the height of a closed right circular cylinder of given surface and maximum volume is equal to the diameter of base.
Solution:
Let h be the height and r be the radius of closed cylinder.
Let S = surface area of cylinder
⇒ S = 2πr + 2πrh ……………(1) (given)
Let V = volume of cylinder = πr2h
= πr2 $$\left[\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right]$$
⇒ V = $$\frac{r}{2}$$ [S – 2πr2]
⇒ $$\frac{d V}{d r}$$ = $$\frac{1}{2}$$ [S – 6πr2]
For MaximaJMinima, $$\frac{d V}{d r}$$ = 0
⇒ S = 6πr2
∴ $$\frac{d^2 V}{d r^2}$$ = $$\frac{1}{2}$$ (- 12πr)
= – 6πr
= – 6π$$\sqrt{\frac{\mathrm{S}}{6 \pi}}$$ < 0 [∵ r > 0]
Thus V is maximise when S = 6πr2
2πr2 + 2πrh = 6πr2 [using (1)]
⇒ 2πrh = 4πr2
⇒ h = 2r
Thus height of cylinder = diameter of closed cylinder.

Question 35.
(i) Show that a closed cylindrical vessel of given volume has the least (total) surface area when its height is twice its radius.
(ii) A closed right circular cylinder has a volume of 2156 cu. cm. What will be the radius of its base so that its total surface area is minimum ? Find the height of the cylinder when its total surface area is minimum. Take π = $$\frac{22}{7}$$. (ISC 2004)
Solution:
(1) Let h be the height and r be the radius of closed cylinder
V = πr2h (given) …………(1)
and Let S = surface area of cylinder = 2πr2 + 2πrh
⇒ S = 2πr2 + 2πr $$\left(\frac{\mathrm{V}}{\pi r^2}\right)$$
= 2πr2 + $$\frac{2 V}{r}$$

∴ S is minimise for r3 = $$\frac{V}{2 \pi}$$
i.e. V = 2πr3
⇒ 2πr3 = πr2h
⇒ h = 2r [using (1)]
i.e. height of cylinder diameter of cylinder.

(ii) Let r be the radius of the base of cylinder
and h be the height of cylinder.
Then volume of cylinder = 2156 cm3
⇒ πr2h = 2156
⇒ h = $$\frac{2156}{\pi r^2}$$ ……….(1)
Let S = Total surface area of cylinder
∴ S = 2πr2 + 2πrh
S = 2πr (r + $$\frac{2156}{\pi r^2}$$) [using (1)]
∴ $$\frac{d \mathrm{~S}}{d r}=2 \pi\left(2 r-\frac{2156}{\pi r^2}\right)$$
For maxima/minima, $$\frac{d S}{d r}$$ = 0
⇒ 2r – $$\frac{2156}{\pi r^2}$$ = 0
⇒ r3 = $$\frac{2156}{2 \pi}=\frac{1078}{22}$$ × 7
= 49 × t = 73
⇒ r = 7 cm
∴ $$\left(\frac{d^2 \mathrm{~S}}{d r^2}\right)_{r=7}=2 \pi\left(2+\frac{4312}{\pi \times 7^3}\right)$$
= 2π $$\left(2+\frac{4312}{22 \times 49}\right)$$
= 12π > 0
Thus, S is minimise for r = 7
Hence the required radius of base of cylinder = 7 cm.

Question 36.
Show that the radius of a closed right circular cylinder of given surface area and maximum volume is equal to half of its height. (ISC 2020)
Solution:
Let r be the radius and h be the height of right circular cylinder.
Let S be given surface area of cylinder.
∴ S = 2πr2 + 2πrh
Let V = volume of cylinder = πr2h
⇒ V = πr2 $$\left(\frac{\mathrm{S}-2 \pi r^2}{2 \pi r}\right)$$
[∵ from (1) ;
h = $$\frac{S-2 \pi r^2}{2 \pi r}$$]
⇒ V = $$\frac{r}{2}$$ (S – 2πr2)
= $$\frac{ S r}{2}$$ – πr3
∴ $$\frac{d V}{d r}$$ = $$\frac{S}{2}$$ – 3πr2
For maxima/minima, $$\frac{d V}{d r}$$ = 0
⇒ $$\frac{S}{2}$$ = 3πr2
⇒ S = 6πr2
⇒ 2πr2 + 2πrh = 6πr2
⇒ 2πrh = πr2
⇒ h = 2r
⇒ r = $$\frac{h}{2}$$
∴ $$\frac{d^2 V}{d x^2}$$ = – 6πr
⇒ $$\frac{d^2 V}{d x^2}$$ < 0
Thus V is maximise when r = $$\frac{h}{2}$$
i.e. V is maximum when radius of right circular cylinder is half of its height.

Question 36 (old).
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere. Let DE = x,
∴ radius of cone = $$\sqrt{r^2-x^2}$$ =BE
and height of cane = r + x
Then volume of cone
⇒ V = $$\frac{1}{3}$$ π (r2 – x2) (r + x)
⇒ $$\frac{d \mathrm{~V}}{d x}=\frac{\pi}{3}$$ [r2 – 2rx – 3x2]
⇒ $$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi}{3}$$ (- 6x – 2r)
For Max./minima, $$\frac{d V}{d x}$$ = 0
⇒ 3x2 + 2rx – r2 = 0
⇒ (x + r) (3x – r) = 0
⇒ x = $$\frac{r}{3}$$ [∵ r ≠ x]
at x = $$\frac{r}{3}$$,
⇒ $$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi}{3}$$ (- 4r) < 0
∴ V is maximum for x = $$\frac{r}{3}$$
∴ Max. volume of cone = $$\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)$$
= $$\frac{32 \pi r^3}{81}$$
Thus Max. value of cone = $$\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)$$
= $$\frac{8}{27}$$ (volume of sphere)
∴ height of cone = x + r
= $$\frac{r}{3}$$ + r = $$\frac{4r}{3}$$
= $$\frac{4 \times 12}{3}$$ = 16 cm then.

[Since given radius of sphere = r = 12 cm]

Question 37.
Find the maximum volume of the cylinder which can be inscribed in a sphere of radius k 3√3 cm. (Leave the answer in terms of it). (ISC 2013)
Solution:
Let h be the height and R be the radius of the cylinder that is inscribed in a sphere of radius r.
In ∆EFA,
R2 = r2 – ($$\frac{h}{2}$$)2 ………..(1)

Let V = volume of under
= πr²h
= πh [r2 – $$\left(\frac{h}{2}\right)^2$$]
∴ $$\frac{d V}{d h}$$ = π $$\left[r^2-\frac{3}{4} h^2\right]$$
∴ $$\frac{d^2 V}{d h^2}$$ = – $$\frac{3}{2}$$ πh
For maxima / minima, $$\frac{d V}{d h}$$ = 0
⇒ r2 – $$\frac{3}{4}$$ h2 = 0
⇒ h = $$\frac{2 r}{\sqrt{3}}$$
∴ $$\left(\frac{d^2 \mathrm{~V}}{d h^2}\right)_{h=\frac{2 r}{\sqrt{3}}}=-\frac{3}{2} \pi \times \frac{2 r}{\sqrt{3}}$$
= – $$\frac{3 \pi r}{\sqrt{3}}$$ < 0
∴ V is maximise h = $$\frac{2 r}{\sqrt{3}}$$
∴ Max. volume = π × $$\frac{2 r}{\sqrt{3}}\left[r^2-\frac{1}{4} \times \frac{4 r^2}{3}\right]$$
= $$\frac{2 r \pi}{\sqrt{3}} \times \frac{2 r^2}{3}=\frac{4 \pi r^3}{3 \sqrt{3}}$$
Given radius of sphere r = 3√3 cm
∴ Max. value = $$\frac{4 \pi}{3 \sqrt{3}} \times(3 \sqrt{3})^3$$
= 108π cm3.

Question 38.
A cone is inscribed in a sphere of radius 12 cm. If the volume of the cone Is maximum, find its height.
Solution:
Let ABC be the cone of largest volume that can be inscribed in the sphere, it is understandable that for maximum value, the axis of the cone must be along the diameter of sphere.
Let DE = x,
radius of cone = $$\sqrt{r^2-x^2}$$ = BE
and height of cane = r + x
Then volume of cone
⇒ V = $$\frac{1}{3}$$ π (r2 – x2)
⇒ $$\frac{d V}{d x}$$ = $$\frac{\pi}{3}$$ [r2 – 2rx – 3x2]
⇒ $$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi}{3}$$ [- 6x – 2r]
For Max./minima, $$\frac{d V}{d x}$$ = 0
⇒ 3x2 + 2rx – r2 = 0
⇒ (x + r) (3x – r) = 0
⇒ x = $$\frac{r}{3}$$ [∵ r ≠ x]
at x = $$\frac{r}{3}$$,
$$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi}{3}$$ (- 4r) < 0
∴ V is maximise for x = $$\frac{r}{3}$$
∴ Max. volume of cone = $$\frac{\pi}{3}\left(r^2-\frac{r^2}{9}\right)\left(r+\frac{r}{3}\right)=\frac{32 \pi r^3}{81}$$
Thus Max. volume of cone = $$\frac{8}{27}\left(\frac{4}{3} \pi r^3\right)$$
= $$\frac{8}{27}$$ (volume of the sphere)
∴ height of cone = x + r
= $$\frac{r}{3}$$ + r
= $$\frac{4 r}{3}=\frac{4 \times 12}{3}$$
= 16 cm then.

[Since given radius of sphere = r = 12 cm].

Question 39.
Show that the altitude of a right circular cone of maximum curved surface area which can be inscribed in a sphere of radius r is $$\frac{4 r}{3}$$.
Solution:
Let PAB be a cone of maximum curved area that can be inscribed in a sphere of radius r.
Now it is obvious that. For maximum volume, axis of cone must lie along diameter of sphere.

∴ PM height of cone = OP + OM = r + x
In right angled ∆OAM,
AM2 = OA2 – OM2
= r2 – x2
Let S = curved surface area of cone
= π (AM) l
= π $$\sqrt{r^2-x^2}\left[\sqrt{(r+x)^2+r^2-x^2}\right]$$
⇒ V = S2 = π2 (r2 – x2) (2r2 + 2rx)
To maximise S it is convenient to maximise S2 = V
∴ $$\frac{d V}{d x}$$ = π2 [2r3 – 4r2x – 6rx2]
and $$\frac{d^2 V}{d x^2}$$ = π2 [- 4r2 – 12rx]
For maxima / minima, $$\frac{d V}{d x}$$ = 0
⇒ 2r3 – 4r2x – 6rx2 = 0
⇒ – 2r (3x2 + 2rx – r2) = 0
⇒ (x + r) (3x – r) = 0
⇒ x = – r, $$\frac{r}{3}$$ [but x > 0]
∴ x = $$\frac{r}{3}$$
∴ $$\left(\frac{d^2 \mathrm{~V}}{d x^2}\right)_{x=\frac{r}{3}}$$
= – 8π2r2 < 0
Thus curved surface area of cone is maximise when x = $$\frac{r}{3}$$
:. altitude of cone = r + x = r + $$\frac{r}{3}$$ = $$\frac{4 r}{3}$$

Question 40.
Show that the height of a right circular cylinder of greatest volume which can be inscribed in a right circular cone of height h and radius r is one-third of the height of the cone and the greatest volume of the cylinder is $$\frac{4}{9}$$ times the volume of the cone.
Solution:
Let PAB be the cone with height OP = h
and OA = OB = r
Let a cylinder of base radius OM’ = ON’ = x
and height = OO’ be inscribed in cone.
Now ∆POB ~ ∆NN’B

Thus, $$\frac{\mathrm{OP}}{\mathrm{NN}^{\prime}}=\frac{\mathrm{OB}}{\mathrm{N}^{\prime} \mathrm{B}}$$
⇒ $$\frac{h}{\mathrm{NN}^{\prime}}=\frac{r}{r-x}$$
NN’ = $$\frac{h(r-x)}{r}$$ = height of cylinder
V = volume of right circular cylinder
= πx2 (NN’)
⇒ V = πx2 $$\frac{h}{r}$$ (r – x)
= $$\frac{\pi h}{r}$$ (rx2 – x3)
∴ $$\frac{d \mathrm{~V}}{d x}=\frac{\pi h}{r}$$ (2rx – 3x2)
and $$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi h}{r}$$ (2r – 6x)
For maxima / minima, $$\frac{d V}{d x}$$ = 0
⇒ $$\frac{\pi h}{r}$$ (2rx – 3x2) = 0
⇒ 2rx = 3x2
⇒ x = $$\frac{2 r}{3}$$
and $$\frac{d^2 V}{d x^2}$$ = $$\frac{\pi h}{r}$$ (2r – 4r)
= – 2πh < 0 [∵ x > 0]
Thus V is maximise when x = $$\frac{2 r}{3}$$
∴ NN’ = height of cylinder
= $$\frac{h}{r}\left[r-\frac{2 r}{3}\right]=\frac{h}{3}$$
Hence volume of right circular cylinder which is inscribed in right circular cone be greatest when height of cylinder = $$\frac{1}{3}$$ height
of cone and greatest volume of cylinder
= πx2$$\frac{h}{3}$$
= $$\pi\left(\frac{2 r}{3}\right)^2 \frac{h}{3}$$
= $$\frac{4}{27}$$ πr2h
= $$\frac{4}{9}$$ ($$\frac{1}{3}$$ πr2h)
= $$\frac{4}{9}$$ (Volume of cone).

Question 41.
The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and $$\frac{x}{3}$$ a sphere is given to be constant, prove that the sum of their volumes is minimum if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
Solution:
S = sum of areas of || piped and sphere
⇒ S = 2 (x × 2x + 2x . $$\frac{x}{3}$$ + x . $$\frac{x}{3}$$) + 4πy2
where y = radius of sphere
⇒ S = 2 (2x2 + $$\frac{2}{3}$$ x2 + $$\frac{x^{2}}{3}$$) + 4πy2
⇒ S = 6x2 + 4πy2 …………..(1)
Let V = volume of ||piped + Volume of sphere
= x (2x) $$\frac{x}{3}$$ + $$\frac{4}{3}$$ πy3

Thus V is minimise when x = 3y
∴ V = $$\frac{4}{3} \pi y^3+\frac{2}{3} x^3$$
= $$\frac{4 \pi}{3}\left(\frac{x}{3}\right)^3+\frac{2}{3} x^3$$
V = $$\frac{4 \pi}{81} x^3+\frac{2}{3} x^3$$
= $$\frac{2}{3} x^3\left(1+\frac{2 \pi}{27}\right)$$.

Question 42.
A manufacturer plans to construct a cylindrical can to hold one cubic metre of oil. If the cost of constructing top and bottom of the can is twice the cost of constructing the side, what are the dimensions ofthe most economical can?
Solution:
Let r be the radius of cylindrical can and h be the height of cylindrical can.
Then 1 = πr2h volume of cylindrical can ………..(1)
Let Rs. p be the cost of constructing the side
then cost of constructing top and bottom of can be Rs. 2p each.
Let C = total cost of constructing the cylindrical can
⇒ C = 2πr2 × 2p + 2πrh × p
⇒ C = 4pπr2 + 2πrp × $$\frac{1}{\pi r^2}$$ [using eqn. (1)]
⇒ C = 4pπr2 + $$\frac{2 p}{r}$$
On differentiating w.r.t. r, we have
$$\frac{d C}{d r}$$ = 8pπr – $$\frac{2 p}{r^{2}}$$ ;
$$\frac{d^2 \mathrm{C}}{d r^2}=8 p \pi+\frac{4 p}{r^3}$$
For maxima / minima,
$$\frac{d C}{d r}$$ = 0
⇒ 8pπr = $$\frac{2 p}{r^{2}}$$
⇒ r3 = $$\frac{1}{4 \pi}$$
⇒ r = $$\left(\frac{1}{4 \pi}\right)^{1 / 3}$$
at r = $$\left(\frac{1}{4 \pi}\right)^{1 / 3}$$ ;
$$\frac{d^2 C}{d r^2}$$ = 8πp + 4p × 4π
= 24pπ > 0
Thus C is maximise for r = $$\left(\frac{1}{4 \pi}\right)^{1 / 3}$$
∴ from (1) ; we have
h = $$\frac{1}{\pi r^2}$$
= $$\frac{1}{\pi\left(\frac{1}{4 \pi}\right)^{2 / 3}}=\frac{(4 \pi)^{2 / 3}}{\pi}$$
⇒ h = $$\left(\frac{16}{\pi}\right)^{1 / 3}$$
Hence, the required dimensions of the most economical can be r = $$\left(\frac{1}{4 \pi}\right)^{1 / 3}$$ metre and
h = $$\left(\frac{16}{\pi}\right)^{1 / 3}$$ metre.

[Since given radius of sphere = r = 12 cm]