Question

The sum of the perimeter 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.

Answer 1

Let side of square be a units and radius of circle be r units.

It is given that 4a + 2πr = k,

Where k is a constant,

⇒ \(r=\frac{k-4a}{2\pi}\) 

Sum of areas, 

A = a² + πr²

Differentiating with respect to a, we get

[ As k = 4a + 2πr given]

⇒ a = 2r

Now, again differentiating equation (i) with respect to a,

∴ For ax = 2r,sum of areas is least.

Hence, sum of areas is least when side of the square is double the radius of the circle.