Question

Consider all rectangles lying in the region
$\left\{(x, y) \in R \times R: 0 \leq x \leq \frac{\pi}{2}\right.$ and $\left.0 \leq y \leq 2 \sin (2 x)\right\}$
and having one side on the $x$-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

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

Answer 1

Answer: (C)

$P(x)=2\left(\frac{\pi}{2}-2 x\right)+4 \sin 2 x$
$=\pi-4 x+4 \sin 2 x$
$\frac{d P}{d x}=-4+8 \cos 2 x$
$\frac{d P}{d x}=0 \Rightarrow x=\frac{\pi}{6}$
$\frac{d^{2} P}{d x^{2}}=-16 \sin \frac{\pi}{3}<0$,
so local maxima exists at $x=\frac{\pi}{6}$
$\therefore $ Length of one side $=2 \sin 2 x=\sqrt{3}$
Length of other side $=\frac{\pi}{6}$
so area $=\sqrt{3} \frac{\pi}{6}=\frac{\pi}{2 \sqrt{3}}$