Question

A rectangle with its sides parallel to the $x$-axis and $y$-axis is inscribed in the region bounded by the curves $y=x^2–4$ and $2y=4–x^2$. The maximum possible area of such a rectangle is closest to the integer

• A. 10
• B. 9
• C. 8
• D. 7

Answer 1

The correct option is B 9

Given:
$y=x^2-4$ and $2y=4-x^2$


$A=(x,\frac{4-x^2}{2})B=(-x,\frac{4-x^2}{2})C=(-x,x^2-4)D=(x,x^2-4)$
Now,
$AB=2xAD=\frac{4-x^2}{2}-(x^2-4)=\frac{12-3x^2}{2}$
The area of the rectangle $=2x\times \frac{12-3x^2}{2}=12x-3x^3$
Maximizing the area, differentiating w.r.t. $x$
$\frac{d\left(area\right)}{dx}=12-9x^2=0\Rightarrow x=\frac{2}{\sqrt{3}}$
$\therefore area=\frac{2}{\sqrt{3}}(12-3\times \frac{4}{3})=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}\approx 9$