Question

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

• A. $\pi r^2$
• B. $r^2$
• C. $\frac{\pi r^2}{4}$
• D. $2r^2$

Answer 1

The correct option is D

$2r^2$

Explanation for the correct option:

${\left(2r\right)}^2=l^2+b^2$

$\Rightarrow 4r^2=l^2+b^2$

$\Rightarrow b=\sqrt{4r^2–l^2}$ ……….$\left(i\right)$

Area $=l\times b=l\times \sqrt{4r^2–l^2}$

$\frac{dA}{dl}=\sqrt{4r^2–l^2}+\left(l\times \frac{1}{2}{\left(4r^2–l^2\right)}^{-\frac{1}{2}}(-2l)\right)$

$\frac{dA}{dl}=\sqrt{4r^2–l^2}-\left(\frac{l^2}{\sqrt{\left(4r^2–l^2\right)}}\right)$

$\frac{dA}{dl}=\left(\frac{4r^2–2l^2}{\sqrt{\left(4r^2–l^2\right)}}\right)$

We know that for critical points, $\frac{dA}{dl}=0$

$\Rightarrow \left(\frac{4r^2–2l^2}{\sqrt{\left(4r^2–l^2\right)}}\right)=0$

$\Rightarrow 4r^2–2l^2=0$

$\Rightarrow 4r^2=2l^2$

$\Rightarrow l=\pm \sqrt{2}r$

From $\left(i\right)$, $b=\sqrt{4r^2–l^2}$

$\therefore b=\sqrt{4r^2–{\left(\sqrt{2}r\right)}^2}=\sqrt{2}r$

So, Area $=l\times b=\sqrt{2}r\times \sqrt{2}r=2r^2$

So, for the maximum area, the length and breadth of the rectangle should be equal. Or, we can say that the Square is the largest rectangle.

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is the correct answer.