# The sum of the perimeters of circle and a square is K, where K is some constant. Prove that the sum of their areas is least when the side of the square is twice the radius of the circle.

Let radius of circle be 'r' and side length of square be 'a' According to questions, $K= 4a+2\pi r$
$\Rightarrow a= \frac{K-2\pi r}{4}---(i)$
Now, combined area $A= a^{2}+\pi r^{2}$
$\Rightarrow A= \left [ \frac{K-2\pi r}{4} \right ]^{2}+\pi r^{2}= \frac{1}{16}\left ( K-2\pi r \right )^{2}+\pi r^{2}$
Differentiating w.r.t r both sides $\frac{dA}{dr}= \frac{1}{4}\pi \left ( K-2\pi r \right )+2\pi r$
Again Diff w.r.t r both sides       $\frac{d^{2}A}{dr^{2}}=\frac{1}{2}\pi ^{2}+2\pi$
for $\frac{dA}{dr}=0$
$\Rightarrow \frac{1}{4}\pi \left ( K-2\pi r \right )+2\pi r= 0\; \; \therefore r= \frac{K}{2\pi +8}$
$\because \frac{d^{2}A}{dr^{2}}]_{at r= \frac{K}{2\pi +8}}= \frac{1}{2}\pi ^{2}+2\pi > 0$
$\therefore A\: is\; least\; at\; r= \frac{K}{2\pi +8}$
Now, $2\pi r+8r= K$
$\Rightarrow 2\pi r+8r= 4a+2\pi r$
$\therefore 2r= a$
Hence the sum of the area of circle and square is least when the side of the square is equal to double the radius of the circle.

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