If rectangles are inscribed in a circle of radius r units. Then the dimensions of the rectangle which has maximum area are

Let \(A B C D\) be the rectangle inscribed in a circle of radius 'r'.
\(\Rightarrow \mathrm{AC}=\mathrm{BD}=2 \mathrm{r}=\) diameter
Let \(x\) and \(y\) be the length and breath of rectangle.
\(\therefore x^{2}+y^{2}=(2 r)^{2} \Rightarrow y=\sqrt{4 r^{2}-x^{2}}\)
Now Area of rectangle \(=\mathrm{A}=\mathrm{xy}\)
\(\therefore A=x \sqrt{4 r^{2}-x^{2}} \)
\( \therefore \frac{d A}{d x}=\sqrt{4 r^{2}-x^{2}}+\frac{x}{2 \sqrt{4 r^{2}-x^{2}}} \times(-2 x)=\)\(\sqrt{4 r^{2}-x^{2}}-\frac{x^{2}}{\sqrt{4 r^{2}-x^{2}}} \)
\( \frac{d A}{d x}=\frac{4 r^{2}-2 x^{2}}{\sqrt{4 r^{2}-x^{2}}}\)
For maximum Area, \(\frac{\mathrm{dA}}{\mathrm{dx}}=0 \Rightarrow 4 \mathrm{r}^{2}-2 \mathrm{x}^{2}=0 \Rightarrow \mathrm{x}=\sqrt{2} \mathrm{r}\)
\(
\frac{\mathrm{d}^{2} \mathrm{~A}}{\mathrm{dx}^{2}}>0 \text { for } \mathrm{x}=\sqrt{2} \mathrm{r}
\)
Therefore area is maximum when \(x=\sqrt{2} r \Rightarrow y=\sqrt{4 r^{2}-2 r^{2}}=\sqrt{2} r\)