Question

Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius $10cm$ is square of side $10\sqrt{2}cm$.

Answer 1

Let $ABCD$ be a rectangle inscribed in a circle of radius $10cm$ with centre at $O$, then $DB=20cm$.
Let $\angle OBA=\theta ,(0<\theta <\frac{\pi }{2})$ and $\theta$ is in radian.
Then, $AB=20\cos \theta$
$AD=20\sin \theta$
Let $p$ be the perimeter of the rectangle $ABCD$, then
$p=2AB+2AD$
$=2(20\cos \theta +20\sin \theta )$
$p=40(\cos \theta +\sin \theta )$ ...$\left(i\right)$

Differentiating (i) w.r.t. $\theta$, we get
$\frac{dp}{d\theta }=40(-\sin \theta +\cos \theta )$ ...$\left(ii\right)$

Again differentiating, we get
$\frac{d^{2}p}{d{\theta }^2}=-40(\cos \theta +\sin \theta )$ ...$\left(iii\right)$

Now, for maximum value $\frac{dp}{d\theta }=0$
$\Rightarrow 40(-\sin \theta +\cos \theta )=0$
$\Rightarrow \cos \theta =\sin \theta$
$\Rightarrow \tan \theta =1$
$\theta ={\tan }^{-1}\left(1\right)$
$\Rightarrow \theta =\frac{\pi }{4}$

Also,

$\left(\begin{array}{c}{\frac{d^{2}p}{d{\theta }^2}}_{\theta =\pi /4}\end{array}\right)$

$=-40\left(\begin{array}{c}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\end{array}\right)$

$=-40\times \frac{2}{\sqrt{2}}$

$=-40\sqrt{2}<0(-ve)$

$\Rightarrow p$ is maximum when $\theta =\frac{\pi }{4}$.

Therefore, $p$ is maximum when
$AB=20\cos \frac{\pi }{4}=20\times \frac{1}{\sqrt{2}}=10\sqrt{2}$
$AD=20\sin \frac{\pi }{4}=20\times \frac{1}{\sqrt{2}}=10\sqrt{2}$
i.e., when adjacent sides are equal and each of them is $10\sqrt{2}cm$.

That is, a rectangle which can be inscribed in a circle of radius 10 cm is a square of side $10\sqrt{2}cm$.