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

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



Now, from figure,
\(\begin{array}{ll}
& A C=\sqrt{a^{2}+b^{2}}=2 r \quad(\because r=\text { Radius }) \\
\Rightarrow & a^{2}+b^{2}=4 r^{2} \\
\Rightarrow & b^{2}=4 r^{2}-a^{2} \\
\Rightarrow & b=\sqrt{4 r^{2}-a^{2}}...(i)
\end{array}\)
Then from Eq. (i), we get
\(A=a \sqrt{4 r^{2}-a^{2}}\)
\(\Rightarrow \quad A^{2}=\left(4 a^{2} r^{2}-a^{4}\right)\)
(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
\(\begin{aligned}
&\frac{d u}{d a}=8 a r^{2}-4 a^{3} \\
&\frac{d^{2} u}{d a^{2}}=8 r^{2}-12 a^{2}
\end{aligned}\)
For max or min,
\(\therefore \quad a=b=2 \sqrt{2}\)
Then, \(\quad \frac{d^{2} y}{d a^{2}}=8(2)^{2}-12(2 \sqrt{2})^{2}\)
\(=32-96 < 0\)
\(\therefore\) Rectangle of maximum area is a square with each side
\(a=b=2 \sqrt{2}\)
Hence, maximum area \(=2 \sqrt{2} \cdot 2 \sqrt{2}=8 \mathrm{sq}\) units