Using integration, find the area of the greatest rectangle that can be inscribed in an ellipse

Using integration, find the area of the greatest rectangle that can be inscribed in an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1.$

Answers (1)

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$
$\Rightarrow y= \frac{b}{a}\sqrt{a^{2}-x^{2}}---(i)$
Let ABCD be the rectangle
Also $AD= 2x,AB= 2y$
Now area, $A= 4xy$
$\Rightarrow A^{2}= 16x^{2}y^{2}$
$\Rightarrow A^{2}= 16x^{2}\, \frac{b^{2}}{a^{2}}\left ( a^{2}-x^{2} \right )$
$\therefore f\left ( x \right )= \frac{16b^{2}}{a^{2}}\left ( a^{2}x^{2} -x^{4}\right )$ where $A^{2}= f\left ( x \right )$
$\Rightarrow {f}'\left ( x \right )= \frac{16b^{2}}{a^{2}}\left ( 2a^{2}x-4x^{3} \right )$
and   $\Rightarrow {f}'\left ( x \right )= \frac{32b^{2}}{a^{2}}\left ( a^{2}-6x^{2} \right )$
For $\Rightarrow {f}'\left ( x \right )= 0,\frac{16b^{2}}{a^{2}}\left ( 2a^{2}x-4x^{3} \right )= 0$
$\Rightarrow x= \frac{a}{\sqrt{2}}\; \; so\: y= \frac{b}{\sqrt{2}}\left [ from(i) \right ]$
$\because {f}''\left ( \frac{a}{\sqrt{2}} \right )= \frac{32b^{2}}{a^{2}}\left [a ^{2}-6\times \frac{a^{2}}{2} \right ]= -64b^{2}< 0$
$\therefore f\left ( x \right )$ is maximum and thus, it is also maximum at $x= \frac{a}{\sqrt{2}}$
Now required area   $=$ or $\left ( OEAFO \right )= 4\int_{0}^{\frac{a}{\sqrt{2}}}y \, dx$
   $\Rightarrow 4\int_{0}^{\frac{a}{\sqrt{2}}}\frac{b}{\sqrt{2}}\, dx$
$\Rightarrow 2\sqrt{2}b\left [ x \right ]^{\frac{a}{\sqrt{2}}}_{0}= 2\sqrt{2}b\left [ \frac{a}{\sqrt{2}}-0 \right ]$
$=$ 2ab sq.units