Question

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x-$axis and vertices $C$ and $D$ lie on the parabola, $y=x^2-1$ below the $x-$axis, is

• A. $\frac{4}{3\sqrt{3}}$
• B. $\frac{4}{3}$
• C. $\frac{2}{3\sqrt{3}}$
• D. $\frac{1}{3\sqrt{3}}$

Answer 1

The correct option is A $\frac{4}{3\sqrt{3}}$


Area $=2a(a^2-1)$
(Say) $A=2a^3-2a$
$\Rightarrow \frac{dA}{da}=6a^2-2$
For maximum and minimum, $\frac{dA}{da}=0$
$\therefore 6a^2-2=0$
$\Rightarrow a=\pm \frac{1}{\sqrt{3}}$
Now, $\frac{d^{2}A}{d{a}^2}=12a$
At $a=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{d^{2}A}{d{a}^2}=4\sqrt{3}>0$
$\therefore a=\frac{1}{\sqrt{3}}$ (rejected)

At $a=-\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{d^{2}A}{d{a}^2}=-4\sqrt{3}<0$
$\therefore$ Maximum area
$=(2a^3-2a)=(-\frac{2}{3\sqrt{3}}+\frac{2}{\sqrt{3}})$
$=\frac{4}{3\sqrt{3}}$ sq. units