Question

A rectangle is inscribed in a semicircle of radius $r$ with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle to get maximum area. Also, find the maximum area.

Answer 1

Let $ABCD$ be the rectangle of sides $x$ and $2y$ inscribed in the seimi-circle with centre $O$ and radius $r$. Let $OA=OB=y$ then $AB=2y$.

Let $\theta$ be the angle, then $x=r\sin \theta ,y=r\cos \theta$
If $A$ is the area of rectangle $ABCD$, then
$A=2yx=2r\cos \theta .r\sin \theta$
$A=r^2\sin 2\theta$, differentiating w.r.t $\theta$
$\therefore \frac{dA}{d\theta }=r^2\cos 2\theta .......\left(i\right)$

For max. or min. $\frac{dA}{d\theta }=0\Rightarrow 2r^2\cos 2\theta =0$
$\cos 2\theta =0=\cos \frac{\pi }{2}\Rightarrow \theta =\frac{\pi }{4}$

Again differentiating equation $\left(i\right)$ w.r.t $\theta$, we get
$\frac{d^{2}A}{d{\theta }^2}=-r^2\sin 2\theta =-4r^2\sin \frac{\pi }{2}=-4r^2$ at $\theta =\frac{\pi }{4}$ which is $-$ve
$\therefore$ For maximum area, $\theta =\frac{\pi }{4}$ and $x=r\sin \theta =r\sin \frac{\pi }{4}=\frac{r}{\sqrt{2}}$
$2y=2r\cos \theta =2r\cos \frac{\pi }{4}=2r\times \frac{1}{\sqrt{2}}=r\sqrt{2}$
Hence, the dimensions of the rectangle are $\frac{r}{\sqrt{2}}$ and $r\sqrt{2}$
and the area $\frac{r}{\sqrt{2}}\times r\sqrt{2}=r^2$ units.