Consider all rectangles lying in the region

$$\left\{(x, y) \epsilon R \times R : 0 \leq x \leq \frac{\pi}{2} and 0 \leq y \leq 2\sin(2x)\right\}$$

and having one side on the $$x$$-axis. The area of the rectangle which has the maximum perimeter among all such rectangles, is

The region is bounded by the curve $$y = 2\sin 2x$$, the $$x$$-axis and the lines $$x = 0$$, $$x = \tfrac{\pi}{2}$$. Every admissible rectangle has its base on the $$x$$-axis, say from $$x = p$$ to $$x = q$$, with $$0 \le p \lt q \le \tfrac{\pi}{2}$$.

Let $$h = 2\min\{\sin 2p,\; \sin 2q\}$$ be the maximum possible height that keeps the whole rectangle below the curve, and let $$w = q-p$$ be its width. The perimeter is therefore

$$P = 2(h+w) = 4\Big(\min\{\sin 2p,\; \sin 2q\} + \tfrac{q-p}{2}\Big).$$

For a fixed left end $$p$$, increasing $$q$$ while $$\sin 2q \ge \sin 2p$$ increases $$w$$ without decreasing $$h$$, hence increases $$P$$. Thus the perimeter is maximised when both endpoints touch the curve at the same height:

$$\sin 2p = \sin 2q.$$

In the interval $$0 \le x \le \tfrac{\pi}{2}$$ the sine function is symmetric about $$x = \tfrac{\pi}{4}$$, so the condition $$\sin 2p = \sin 2q$$ forces

$$p = \tfrac{\pi}{4}-t, \qquad q = \tfrac{\pi}{4}+t,$$ with $$0 \le t \le \tfrac{\pi}{4}.$

The rectangle then has
width $$w = q-p = 2t,$$
height $$h = 2$$\sin$$ 2p = 2$$\sin$$\!\bigl(\tfrac{$$\pi$$}{2}-2t\bigr) = 2$$\cos$$ 2t.$$

Hence the perimeter becomes

$$P(t) = 2(h+w) = 4\bigl($$\cos$$ 2t + t\bigr).$$

To maximise $$P(t)$$, differentiate:

$$P'(t) = 4(-2$$\sin$$ 2t + 1).$$

Set $$P'(t)=0$$:

$$-2$$\sin$$ 2t + 1 = 0 \;\;\Longrightarrow\;\; $$\sin$$ 2t = \tfrac12.$$

Within $$0 \le t \le \tfrac{$$\pi$$}{4}$$ this gives $$2t = \tfrac{$$\pi$$}{6}\; \Longrightarrow\; t = \tfrac{$$\pi$$}{12}.$$

Second derivative: $$P''(t) = 4(-4$$\cos$$ 2t) = -16$$\cos$$ 2t.$$ At $$t = \tfrac{$$\pi$$}{12},\; $$\cos$$ 2t = $$\cos$$ \tfrac{$$\pi$$}{6} = \tfrac{$$\sqrt$$3}{2} \gt 0$$, so $$P''(t) \lt 0$$, confirming a maximum.

For $$t = \tfrac{$$\pi$$}{12}$$,
width $$w = 2t = \tfrac{$$\pi$$}{6},$$
height $$h = 2$$\cos$$ 2t = 2$$\cos$$ \tfrac{$$\pi$$}{6} = 2$$\left$$(\tfrac{$$\sqrt$$3}{2}$$\right$$) = $$\sqrt$$3.$$

Therefore the maximum perimeter rectangle has area

$$A = wh = $$\left$$(\tfrac{$$\pi$$}{6}$$\right$$)\!($$\sqrt$$3) = \tfrac{$$\pi$$$$\sqrt$$3}{6} = $$\frac{\pi}{2\sqrt3}$$.$$

Thus the required area is $$\boxed{\dfrac{$$\pi$$}{2$$\sqrt$$3}}.$$ Option C is correct.