Question

The maximum area of a rectangle that can be inscribed in a circle of radius $2$ units is

• A. 8 $\pi$ sq. units
• B. 4 sq. units
• C. 5 sq. units
• D. 8 sq. units

Answer 1

Answer: (D)

Let length of the rectangle $=a$ and breadth $=b$
Then, area of rectangle,
$A=a\cdot b...(i)$

Now, from figure,
$AC=\sqrt{a^2+b^2}=2r(\because r=$ Radius $)$
$\Rightarrow a^2+b^2=4r^2$
$\Rightarrow b^2=4r^2-a^2$
$\Rightarrow b=\sqrt{4r^2-a^2}\dots (ii)$
Then from Eq. (i). we get
$A=a\sqrt{4r^2-a^2}$
$\Rightarrow A^2=\left(4a^2{r}^2-a^4\right)$
(squaring on both sides) Let
$u=4a^2{r}^2-a^4...(iii)$
So, $A^2$ is max or min according as $u$ is max or min. Now, differentiate Eq. (iii) two times w.r.t. $a$, we get
$\frac{du}{da}=8a{r}^2-4a^3$
$\frac{d^{2}u}{d{a}^2}=8r^2-12a^2$
For max or min,
Put $\frac{du}{da}=8a{r}^2-4a^3=0$
$\Rightarrow 4a\left(2r^2-a^2\right)=0$
$\Rightarrow a=\sqrt{2}\cdot r(\because a\ne 0)$
$\therefore b=\sqrt{4r^2-(\sqrt{2}r{)}^2}$
$=\sqrt{4r^2-2r^2}=\sqrt{2}\cdot r[$ from Eq. (ii)]
Given, radius $(r)=2$ units
$\therefore a=b=2\sqrt{2}$
Then, $\frac{d^{2}y}{d{a}^2}=8(2{)}^2-12(2\sqrt{2}{)}^2$
$=32-96<0$(max)
$\therefore$ Rectangle of maximum area is a square with each side
$a=b=2\sqrt{2}$
Hence, maximum area $=2\sqrt{2}\cdot 2\sqrt{2}=8sq$ units