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.