Question

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

Answer 1

Let $ABCD$ be a rectangle inscribed in a given circle with centre $O$ and radius $a$.
Let $AB=2x$ and $BC=2y$.

Applying Pythagoras theorem in $△OAM$, we obtain
$O{A}^2=A{M}^2+O{M}^2$
$\Rightarrow a^2=x^2+y^2\Rightarrow y=\sqrt{a^2-x^2}...\left(1\right)$

Let $A$ be the area of the rectangle $ABCD$. Then, $A=4xy=4x\sqrt{a^2-x^2}$
Differentiate w.r.t $x$,
$\frac{dA}{dx}=4[\sqrt{a^2-x^2}-\frac{x^2}{\sqrt{a^2-x^2}}]=4\left[\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\right]$

The critical points of $A$ are given by $\frac{dA}{dx}=0$
$\frac{dA}{dx}=0\Rightarrow 4\left[\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\right]=0$
$\Rightarrow a^2-2x^2=0\Rightarrow x=\frac{a}{\sqrt{2}}$

Now, $\frac{dA}{dx}=4\left[\frac{a^2-2x^2}{\sqrt{a^2-x^2}}\right]$
$\frac{d^{2}A}{d{x}^2}=4\left[\frac{(a^2-x^2{)}^{1/2}(-4x)-(a^2-2x^2)\cdot \frac{1}{2}\cdot (a^2-x^2{)}^{-1/2}(-2x)}{a^2-x^2}\right]$
$=4\left[\frac{-4x(a^2-x^2{)}^{1/2}+x(a^2-2x^2\left)\right(a^2-x^2{)}^{-1/2}}{a^2-x^2}\right]$
$\frac{d^{2}A}{d{x}^2}{|}_{x=a/\sqrt{2}}=-16<0$

Thus, $A$ is maximum when $x=\frac{a}{\sqrt{2}}$.
Putting $x=\frac{a}{\sqrt{2}}$ in (1), we get $y=\frac{a}{\sqrt{2}}$
$\Rightarrow 2x=2y=a\sqrt{2}$
$\Rightarrow AB=BC\Rightarrow ABCD$ is a square.