The sum of the perimeter of a circle and square is $k$, where $k$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Let $r$ be the radius of the circle and $x$ be the side of the square, then $2\pi r + 4x = k$. …(i)

Let $S$ be the sum of areas of the circle and the square, then
$S = \pi {r^2} + {x^2} = \pi {r^2} + {\left( {\cfrac{{k - 2\pi r}}{4}} \right)^2}$ (from (i)$x = \cfrac{{k - 2\pi r}}{4}$) $= \pi {r^2} + \cfrac{{{k^2}}}{{16}} + \cfrac{{{\pi ^2}{r^2}}}{4} - \cfrac{{k\pi r}}{4}$.

…(ii) Differentiating (ii) w.r.t.$r$, we get $\cfrac{{dS}}{{dr}} = 2\pi r + \cfrac{{2{\pi ^2}r}}{4} - \cfrac{{k\pi }}{4}$ …(iii) For maximum/minimum $S,\cfrac{{dS}}{{dr}} = 0$

$\Rightarrow 2\pi r + \cfrac{{2{\pi ^2}r}}{4} - \cfrac{{k\pi }}{4} = 0 \Rightarrow r\left( {2\pi + \cfrac{{{\pi ^2}}}{2}} \right) = \cfrac{{k\pi }}{4}$
$\Rightarrow r = \cfrac{{2k\pi }}{{4(4\pi + {\pi ^2})}} = \cfrac{k}{{2(4 + \pi )}}$

Differentiating (iii) w.r.t. $r$, we get $\cfrac{{{d^2}S}}{{d{r^2}}} = 2\pi + \cfrac{{{\pi ^2}}}{2} > 0\forall r.$

So, ${\left( {\cfrac{{{d^2}S}}{{d{r^2}}}} \right)_{r = \cfrac{k}{{2(4 + \pi )}}}} > 0 \Rightarrow S$ is minimum at $r = \cfrac{k}{{2(4 + \pi )}}$
$\Rightarrow x = \cfrac{1}{4}\left\{ {k - \cfrac{{2\pi }}{2}\left( {\cfrac{k}{{4 + \pi }}} \right)} \right\}$ (from (i))

$= \cfrac{{4k}}{{4(4 + \pi )}} = 2\left[ {\cfrac{{2k}}{{4(4 + \pi )}}} \right] = 2\left[ {\cfrac{k}{{2(4 + \pi )}}} \right] = 2(r)$

Hence, $S$ is least when side of the square is double the radius of the circle.