Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Let $ABCD$ be a rectangle inscribed in the given circle of radius $r$ having centre at $O$.

Let one side of the rectangle be $x$, then the other side of the rectangle
$= \sqrt {{{(2r)}^2} - {x^2}} = \sqrt {4{r^2} - {x^2}}$ (; an angle in the semicircle) Let $A$ be the corresponding area of the rectangle, then

$A = x\sqrt {4{r^2} - {x^2}} ,0 < x < 2r$
$\Rightarrow \cfrac{{dA}}{{dx}} = \cfrac{{x( - 2x)}}{{2\sqrt {4{r^2} - {x^2}} }} + \sqrt {4{r^2} - {x^2}} = \cfrac{{2(2{r^2} - {x^2})}}{{\sqrt {4{r^2} - {x^2}} }}$

To find the maximum or minimum area,
$\cfrac{{dA}}{{dx}} = 0 \Rightarrow \cfrac{{2{r^2} - {x^2}}}{{\sqrt {4{r^2} - {x^2}} }} = 0 \Rightarrow x = \sqrt 2 r$

Now, $\cfrac{{{d^2}A}}{{d{x^2}}} = \cfrac{{\sqrt {4{r^2} - {x^2}} ( - 4x) - (4{r^2} - 2{x^2})\cfrac{{1 \times ( - 2x)}}{{2\sqrt {4{r^2} - {x^2}} }}}}{{(4{r^2} - {x^2})}}$

$= \cfrac{{(4{r^2} - {x^2})( - 4x) + (4{r^2} - 2{x^2})x}}{{{{(4{r^2} - {x^2})}^{3/2}}}} = \cfrac{{ - 12{r^2}x + 2{x^3}}}{{{{(4{r^2} - {x^2})}^{3/2}}}}$

${\left( {\cfrac{{{d^2}A}}{{d{x^2}}}} \right)_{x = \sqrt 2 r}} = \cfrac{{ - 12{r^2}(\sqrt 2 r) + 2{{(\sqrt 2 r)}^3}}}{{{{(2{r^2})}^{3/2}}}}$

$= \cfrac{{4\sqrt 2 {r^3} - \sqrt 2 {r^3}}}{{2\sqrt 2 {r^3}}} = 2 - 6 < 0$
Therefore area is maximum at $x = \sqrt 2 r$

Therefore the length of rectangle is $\sqrt 2 r$ and width of rectangle is $\sqrt {4{r^2} - 2{r^2}} = \sqrt 2 r$

Hence, of all the rectangles the square of side $\sqrt 2 r$ units has the maximum area.