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

$\text{Region: } 0 \leq x \leq \dfrac{\pi}{2},\quad 0 \leq y \leq 2\sin(2x)$ $\text{Rectangle: one side on the } x\text{-axis, lying inside the region}$ $\text{To find: area of the rectangle of maximum perimeter}$ $\text{Plan: base } [p,q] \text{ on } x\text{-axis, height } h \leq \min_{[p,q]} 2\sin(2x)$ $\text{By symmetry about } x = \pi/4:\ p = \tfrac{\pi}{4}-t,\ q = \tfrac{\pi}{4}+t$ $\text{maximize perimeter, then read the area}$ $y = 2\sin(2x) \text{ peaks at } x = \pi/4 \text{ and is symmetric about it}$ $\text{perimeter is largest when the base is centred at } \pi/4$ $p = \tfrac{\pi}{4}-t,\quad q = \tfrac{\pi}{4}+t,\qquad t \in \left[0, \tfrac{\pi}{4}\right]$ $\text{height } h = 2\cos 2t,\qquad \text{width} = 2t$ $P(t) = 2\bigl(2t + 2\cos 2t\bigr) = 4t + 4\cos 2t$ $P'(t) = 4 - 8\sin 2t = 0 \Rightarrow \sin 2t = \tfrac{1}{2}$ $2t = \tfrac{\pi}{6} \Rightarrow t = \tfrac{\pi}{12}$ $P''(t) = -16\cos 2t < 0 \Rightarrow \text{max at } t = \tfrac{\pi}{12}$ $\text{width} = 2t = \tfrac{\pi}{6},\qquad \text{height} = 2\cos\tfrac{\pi}{6} = \sqrt{3}$ $\text{Area} = \dfrac{\pi}{6}\cdot\sqrt{3} = \dfrac{\pi\sqrt{3}}{6} = \dfrac{\pi}{2\sqrt{3}}$ $\boxed{\text{option (3)}}$