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.


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