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Given the sum of the perimeters of a square and a circle show that the sum of their areas is least when the side of the square is equal to the diameter of a circle.

Given the sum of the perimeters of a square and a circle show that the sum of their areas is least when the side of the square is equal to the diameter of a circle.

Grade:Upto college level

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
6 years ago
Ans:
Let side of square to be ‘a’ & radius of circle to be ‘r’
Given:
P = 4a + 2\pi r
Sum of their area:
A = a^{2}+\pi r^{2}
r = \frac{P-4a}{2\pi }
A = a^{2}+\pi (\frac{P-4a}{2\pi })^{2}
\frac{\partial A}{\partial a} = 2a+2\pi (\frac{P-4a}{2\pi })(\frac{-4}{2\pi }) = 0
\pi a = P - 4a
P = \pi a + 4a
\pi a+4a = 4a+2\pi r
a = 2r
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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