Let length of the rectangle $=a$ and breadth $=b$
Then, area of rectangle,
$A=a \cdot b\,\,\,\,\,\,\,...(i)$
Now, from figure,
$ A C=\sqrt{a^{2}+b^{2}}=2 r \,\,\,\,\,(\because r= $ Radius $) $
$\Rightarrow a^{2}+b^{2}=4 r^{2} $
$\Rightarrow b^{2}=4 r^{2}-a^{2} $
$\Rightarrow b=\sqrt{4 r^{2}-a^{2}} \,\,\,\, \ldots (ii) $
Then from Eq. (i). we get
$A=a \sqrt{4 r^{2}-a^{2}} $
$\Rightarrow A^{2}=\left(4 a^{2} r^{2}-a^{4}\right)$
(squaring on both sides) Let
$u=4 a^{2} r^{2}-a^{4}\,\,\,\,\,\,\,\,\,...(iii)$
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{d u}{d a}=8 a r^{2}-4 a^{3} $
$\frac{d^{2} u}{d a^{2}}=8 r^{2}-12 a^{2}$
For max or min,
Put $ \frac{d u}{d a}=8 a r^{2}-4 a^{3}=0 $
$\Rightarrow 4 a\left(2 r^{2}-a^{2}\right)=0 $
$\Rightarrow a=\sqrt{2} \cdot r \,\,\,\,\,\,\,\, (\because a \neq 0)$
$\therefore b=\sqrt{4 r^{2}-(\sqrt{2} r)^{2}}$
$=\sqrt{4 r^{2}-2 r^{2}}=\sqrt{2} \cdot r \,\,\,\,\,[$ 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 $=2 \sqrt{2} \cdot 2 \sqrt{2}=8\, sq$ units
