Question

Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius $r$.

Answer 1

Let us take the circle with centre $(0,0)$ and radius $r$ and $PQRS$ be the rectangle inscribed in the circle.
Let $x=r\cos \theta ;y=r\sin \theta$
The dimensions of the rectangle are
$2x=2r\cos \theta ;2y=2r\sin \theta$
Area at rectangle $A=4r^2\sin \theta \cos \theta =2r^2\sin 2\theta$
$\Rightarrow A\left(\theta \right)=2r^2\sin 2\theta$
$\Rightarrow A^{\prime }\left(\theta \right)=4r^2\cos 2\theta$
$\Rightarrow A"\left(\theta \right)=-8r^2\sin 2\theta$
$\Rightarrow A^{\prime }\left(\theta \right)=0$
$\Rightarrow 4r^2\cos 2\theta =0$
$\Rightarrow \cos 2\theta =0\Rightarrow 2\theta =\frac{\pi }{2}\Rightarrow \frac{\pi }{4}$
Now $A^{\prime \prime }\left(\frac{\pi }{4}\right)=-8r^2\times 1=-8r^2<0$
A is largest when $\theta =\frac{\pi }{4}$
When $\theta =\frac{\pi }{4}$
$2x=2r\cos \times \frac{\pi }{4}=2r\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2r}$
$2y=2r\sin \frac{\pi }{4}=2r\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2r}$
$\therefore$ the dimensions of the rectangle are $\sqrt{2}r$ and $\sqrt{2}r$.
(The rectangle is also square)