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

= πr^{2} + a^{2} = πr^{2} +

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.

Rate this question :

If the sum of theMathematics - Board Papers

A metal box with Mathematics - Board Papers

Show that aRD Sharma - Volume 1

Find the local maMathematics - Board Papers

Prove that the seMathematics - Board Papers

Prove that the raMathematics - Board Papers

Prove that the leMathematics - Board Papers