A rectangle is inscribed in a semi-circle

Solution:

Let the dimensions of the rectangle be $x$ and $y$. Then,

$\frac{x^{2}}{4}+y^{2}=r^{2}$

$\Rightarrow x^{2}+4 y^{2}=4 r^{2}$

$\Rightarrow x^{2}=4\left(r^{2}-y^{2}\right)$                .....(1)

Area of rectangle $=x y$

$\Rightarrow A=x y$

Squaring both sides, we get

$\Rightarrow A^{2}=x^{2} y^{2}$

$\Rightarrow Z=4 y^{2}\left(r^{2}-y^{2}\right)$          [From eq. (1)]

$\Rightarrow \frac{d Z}{d y}=8 y r^{2}-16 y^{3}$

For the maximum or minimum values of $Z$, we must have

$\frac{d Z}{d y}=0$

$\Rightarrow 8 y r^{2}-16 y^{3}=0$

$\Rightarrow 8 r^{2}=16 y^{2}$

$\Rightarrow y^{2}=\frac{r^{2}}{2}$

$\Rightarrow y=\frac{r}{\sqrt{2}}$

Substituting the value of $y$ in eq. (1), we get

$\Rightarrow x^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right)$

$\Rightarrow x^{2}=4\left(r^{2}-\frac{r^{2}}{2}\right)$

$\Rightarrow x^{2}=4\left(\frac{r^{2}}{2}\right)$

$\Rightarrow x^{2}=2 r^{2}$

$\Rightarrow x=r \sqrt{2}$

Now,

$\frac{d^{2} Z}{d y^{2}}=8 r^{2}-48 y^{2}$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=8 r^{2}-48\left(\frac{r^{2}}{2}\right)$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=-16 r^{2}<0$

So, the area is maximum when $x=r \sqrt{2}$ and $y=\frac{r}{\sqrt{2}}$.

Area $=x y$

$\Rightarrow A=r \sqrt{2} \times \frac{r}{\sqrt{2}}$

$\Rightarrow A=r^{2}$