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 +


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.

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