Question

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.

Accepted Answer

Let $x$ be the side of square and $y$ be the radius of circle.

Let $P$ be the sum of their perimeter.

Then, $P = 4 x + 2 πy . . . . . . . . . . . . . . . . . (1)$

Let, $A =$ combine area of square and circle

$\Rightarrow A = x^{2} + {πy}^{2}$

$A = x^{2} + π {(\frac{P - 4 x}{2 π})}^{2}$ [using $(1)$ ]

$A = x^{2} + \frac{1}{4 π} {(P - 4 x)}^{2}$

On differentiating both the sides w.r.t. $x$ , we get

$\frac{d A}{d x} = 2 x + \frac{1}{2 π} (P - 4 x) (- 4)$

and $\frac{d^{2} A}{d x^{2}} = 2 + \frac{16}{2 π}$

Now, For maxima/minima, $\frac{d A}{d x} = 0$

$\Rightarrow 2 x - \frac{2}{π} (P - 4 x) = 0 \Rightarrow 2 π x - 2 P + 8 x = 0 \Rightarrow x = \frac{2 P}{2 π + 8} = \frac{P}{π + 4} (∵ \frac{P}{π + 4} > 0)$

Now, at $x = \frac{P}{π + 4}$ ;

$(\frac{d^{2} A}{d x^{2}}) = 2 + \frac{16}{2 π} (∵ 2 + \frac{16}{2 π} > 0)$

Thus, $x = \frac{P}{π + 4}$ is the point of minima.

Therefore, area $A$ is minimum when $x = \frac{P}{π + 4}$ .

$∴$ from $(1)$ ; $2 π y = P - \frac{4 P}{π + 4}$

$\Rightarrow 2 π y = \frac{π P}{π + 4} \Rightarrow y = \frac{P}{2 (π + 4)} = \frac{x}{2} \Rightarrow x = 2 y$

i.e. side of square $=$ diameter of circle.

Hence, $A$ is minimum when side of square is equal to the diameter of the circle.

M L Aggarwal Solutions for Chapter: Applications of Derivatives, Exercise 8: EXERCISE 7.8

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Applications of Derivatives, Exercise 8: EXERCISE 7.8

Questions from M L Aggarwal Solutions for Chapter: Applications of Derivatives, Exercise 8: EXERCISE 7.8 with Hints & Solutions

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