Interpret the region and the rectangle

The region is bounded by $0 \le x \le \frac{\pi}{2}, \qquad 0 \le y \le 2\sin(2x).$

Since the rectangle has one side on the $x$-axis and lies symmetrically under the curve, take the top corners of the rectangle at $x=a \quad \text{and} \quad x=\frac{\pi}{2}-a,$ because $2\sin(2x)=2\sin\bigl(2(\tfrac{\pi}{2}-x)\bigr)=2\sin(\pi-2x)=2\sin(2x),$ so the curve is symmetric about $x=\frac{\pi}{4}$.

Then:

• height of rectangle = $2\sin(2a)$,
• width of rectangle = $\frac{\pi}{2}-2a$.

Perimeter of the rectangle

The perimeter is

Maximize the perimeter

Differentiate: $P'(a)=-4+8\cos(2a).$

Set $P'(a)=0$: $-4+8\cos(2a)=0$ $\cos(2a)=\frac12.$

Since $0\le a\le \frac{\pi}{4}$, we get $2a=\frac{\pi}{3} \implies a=\frac{\pi}{6}.$

Second derivative: $P''(a)=-16\sin(2a).$ At $a=\frac{\pi}{6}$, $P''\left(\frac{\pi}{6}\right)=-16\sin\left(\frac{\pi}{3}\right)<0,$ so this gives maximum perimeter.

Find the area of this rectangle

Width: $\frac{\pi}{2}-2\cdot\frac{\pi}{6}=\frac{\pi}{6}.$

Height: $2\sin\left(2\cdot\frac{\pi}{6}\right)=2\sin\left(\frac{\pi}{3}\right)=2\cdot\frac{\sqrt3}{2}=\sqrt3.$

Therefore area is A=\frac{\pi}{6}\cdot \sqrt3=\frac{\pi\sqrt3}{6}=rac{\pi}{2\sqrt3}.

Match with options

$\frac{\pi}{2\sqrt3}$ corresponds to Option C.

Compare with stored answer

Stored correct answer: C

My derived answer: C

So they agree.