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.

Question

Philoid · Accepted Answer

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

Then, 2πr + 4a = k (where k is constant)

The sum of the areas of the circle and square (A) is

= πr² + a² = πr² +

Now,

8r = k - 2πr

⇒ (8 + 2π)r = k

⇒ r =

Now, > 0

Therefore, When r = , > 0

⇒ The sum of the area is least when r =

So, when r =

Then a =

Therefore, it is proved that the sum of their areas is least when the side of the square is double the radius of the circle.