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

Jitender Singh IIT Delhi
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