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Q.
The area of greatest rectangle that can be inscribed in an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, is
Application of Derivatives
Solution:
Let $A B C D$ be the rectangle of maximum area with sides $A B=2 x$ and $B C=2 y$, where $C(x, y)$ is a point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as shown in the figure. The area $A$ of the rectangle is $4 x y$ i.e., $A=4 x y$ which gives $A^2=16 x^2 y^2=s$ (say)
Therefore, $s=16 x^2\left(1-\frac{x^2}{a^2}\right) \cdot b^2=\frac{16 b^2}{a^2}\left(a^2 x^2-x^4\right)$
$\Rightarrow \frac{d s}{d x}=\frac{16 b^2}{a^2} \cdot\left[2 a^2 x-4 x^3\right]$
Again, $\frac{d s}{d x}=0 \Rightarrow x=\frac{a}{\sqrt{2}}$ and $y=\frac{b}{\sqrt{2}}$
Now, $\frac{d^2 s}{d x^2}=\frac{16 b^2}{a^2}\left[2 a^2-12 x^2\right]$
At $x=\frac{a}{\sqrt{2}}, \frac{d^2 s}{d x^2}=\frac{16 b^2}{a^2}\left[2 a^2-6 a^2\right]$
$=\frac{16 b^2}{a^2}\left(-4 a^2\right)<0$
Thus at $x=\frac{a}{\sqrt{2}}, y=\frac{b}{\sqrt{2}}, s$ is maximum and hence the area $A$ is maximum.
Maximum area $=4 \cdot x \cdot y \cdot=4 \cdot \frac{a}{\sqrt{2}} \cdot \frac{b}{\sqrt{2}}=2 a b$ sq units.
