Question

Consider all rectangles lying in the region
$\{(x,y)\in R\times R:0\le x\le \frac{\pi }{2}$ and $0\le y\le 2\sin (2x)\}$
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)

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