The maximum area of the rectangle that can be inscribed in a circle of radius \(r\) is

Refer figure


\(
\begin{aligned}
& \mathrm{B}=(\mathrm{r} \cos \theta, \mathrm{r} \sin \theta) \\
& \therefore \mathrm{Ab} 2 \mathrm{r} \cos \theta \text { and } \mathrm{BC}=2 \mathrm{r} \sin \theta \\
& \text { Area }(\mathrm{ABCD})=\mathrm{AB} \times \mathrm{BC} \\
& \therefore \mathrm{f}(\theta)=(2 \mathrm{r} \cos \theta)(2 \mathrm{r} \sin \theta)=2 \mathrm{r}^2 \sin 2 \theta \\
& \mathrm{f}^{\prime}(\theta)=4 \mathrm{r}^2 \cos 2 \theta \text { and when } \mathrm{f}^{\prime}(\theta)=0 \text {, we write } \\
& \cos 2 \theta=0 \Rightarrow \cos 2 \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4} \\
& \mathrm{f}^{\prime \prime}(\theta)=-8 \mathrm{r}^2 \sin 2 \theta \text { and } \mathrm{f}^{\prime \prime}(\theta)_{\left[\theta=\frac{\pi}{4}\right]}=-8 \mathrm{r}^2 < 0
\end{aligned}
\)
Thus, area of required rectangle is maximum when \(\theta=\frac{\pi}{4}\)
\(
\begin{aligned}
& \therefore \mathrm{AB}=2 \mathrm{r} \cos \frac{\pi}{4}=2 \mathrm{r}\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2} \mathrm{r} \text { and } \\
& \mathrm{BC}=2 \mathrm{r} \sin \frac{\pi}{4}=2 \mathrm{r}\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2} \mathrm{r}
\end{aligned}
\)
\(\therefore\) Maximum Area \((\mathrm{ABCD})==(\sqrt{2} \mathrm{r})(\sqrt{2} \mathrm{r})=2 \mathrm{r}^2\)