Question

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

Answer 1

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

Let $\angle CEB=\theta$
Then, $EC=2r,EB=2rcos\theta$ and $BC=2rsin\theta$
Let A be the area of rectangle BCDE
Then, $A=EB\times BC=4r^{2}sin\theta cos\theta =2r^{2}sin2\theta$
Thus, $A=2r^{2}sin2\theta$
where, r is constant.
$\therefore \frac{dA}{d\theta }=4r^{2}cos2\theta ,\frac{d^{2}A}{d{\theta }^2}=-8r^{2}sin2\theta$
Now, put $\frac{dA}{d\theta }=0$
$\Rightarrow 4r^{2}cos2\theta =0\Rightarrow 2\theta =\frac{\pi }{2}i.e.,\theta =\frac{\pi }{4}$
At $\theta =\frac{\pi }{4},{\left(\frac{d^{2}A}{d{\theta }^2}\right)}_{\theta =\frac{\pi }{4}}=-8r^{2}sin\frac{\pi }{2}=-8r^2<0$
$\therefore \theta =\frac{\pi }{4}$ is a point of maxima.
Thus, area is maximum when $\theta =\frac{\pi }{4}$
In this case, $EB=2rcos\frac{\pi }{4}=r\sqrt{2}$ and $BC=2rsin\frac{\pi }{4}=r\sqrt{2}$
Thus, EB=BC and therefore, BCDE is a square.