#### Question

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

#### Solution

\[\text { Let the length of a side of the square and radius of the circle be x and r, respectively .} \]

\[\text { It is given that the sum of the perimeters of square and circle is constant .} \]

\[ \Rightarrow 4x + 2\pi r = K \left( \text { Where K is some constant } \right)\]

\[ \Rightarrow x = \frac{\left( K - 2\pi r \right)}{4} . . . \left( 1 \right)\]

\[\text { Now,} \]

\[A = x^2 + \pi r^2 \]

\[ \Rightarrow A = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2 \left[ \text { From eq } . \left( 1 \right) \right]\]

\[ \Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right)^2}{16} + \pi r^2 \]

\[ \Rightarrow \frac{dA}{dr} = \frac{2\left( K - 2\pi r \right) - 2\pi}{16} + 2\pi r\]

\[ \Rightarrow \frac{dA}{dr} = \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r\]

\[ \Rightarrow \frac{\left( K - 2\pi r \right) - \pi}{4} + 2\pi r = 0\]

\[ \Rightarrow \frac{\left( K - 2\pi r \right)\pi}{4} = 2\pi r\]

\[ \Rightarrow K - 2\pi r = 8r . . . \left( 2 \right)\]

\[\frac{d^2 A}{d x^2} = \frac{\pi^2}{2} + 2\pi > 0\]

\[\text { So, the sum of the areas, A is least when }K - 2\pi r = 8r . \]

\[\text { From eqs }. \left( 1 \right) \text { and }\left( 2 \right), \text { we get}\]

\[x = \frac{\left( K - 2\pi r \right)}{4}\]

\[ \Rightarrow x = \frac{8r}{4}\]

\[ \Rightarrow x = 2r\]

\[ \therefore\text { Side of the square = Diameter of the circle }\]