Question

The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer 1

Let $r$ be the radius of the circle and $a$ be the side of the square.
Then, we have:
$2πr+4a=k$(where $k$ is constant)
$⇒a=\frac{k-2πr}{4}$
The sum of the areas of the circle and the square $(A)$ is given by,
$A=πr^2+a^2=πr^2+\frac{(k-2πr)^2}{16}$
$∴\frac{dA}{dr}=2πr+\frac{2(k-2πr)(-2π)}{16}=2πr-\frac{π(k-2πr)}{4}$
Now,$\frac{dA}{dr}=0$
$⇒2πr=\frac{π(k-2πr)}{4}$
$8r=k-2πr$
$⇒(8+2π)r=k$
$⇒r=\frac{k}{8+2π}=\frac{k}{2(4+π)}$
Now,$\frac{d^2A}{dr^2}=2π+\frac{π^2}{2}>0$
∴When $r=\frac{k}{2(4π)},\frac{d^2A}{dr^2}>0$
∴ The sum of the areas is least when $r=\frac{k}{2(4π)}.$
When $r=\frac{k}{2(4π)},a=\frac{k-2π[\frac{k}{2(4π)}]}{4}$
$=\frac{k(4π)\,π\,\,k}{4\, 4(π)\,4}=\frac{4k}{4(π)4}=\frac{k}{π}=2r$
Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.