Consider all rectangles lying in the region l | vedclass

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(C) Let the rectangle be symmetric about the line $x = \frac{\pi}{4}$. Let the width of the rectangle be $2\alpha$,where the $x$-coordinates of the ve

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$\frac{\pi}{2 \sqrt{3}}$

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(C) Let the rectangle be symmetric about the line $x = \frac{\pi}{4}$. Let the width of the rectangle be $2\alpha$,where the $x$-coordinates of the vertical sides are $\frac{\pi}{4} - \alpha$ and $\frac{\pi}{4} + \alpha$. The height of the rectangle is $y = 2 \sin(2(\frac{\pi}{4} + \alpha)) = 2 \sin(\frac{\pi}{2} + 2\alpha) = 2 \cos(2\alpha)$. The perimeter $P$ of the rectangle is given by $P = 2(	ext{width} + 	ext{height}) = 2(2\alpha + 2 \cos(2\alpha)) = 4(\alpha + \cos(2\alpha))$. To find the maximum perimeter,we differentiate $P$ with respect to $\alpha$: $\frac{dP}{d\alpha} = 4(1 - 2 \sin(2\alpha)) = 0$. This gives $\sin(2\alpha) = \frac{1}{2}$,so $2\alpha = \frac{\pi}{6}$ (since $0 Checking the second derivative: $\frac{d^2P}{d\alpha^2} = -8 \cos(2\alpha)$. At $\alpha = \frac{\pi}{12}$,$\frac{d^2P}{d\alpha^2} = -8 \cos(\frac{\pi}{6}) = -4\sqrt{3} The area of this rectangle is $	ext{Area} = 	ext{width} 	imes 	ext{height} = (2\alpha) 	imes (2 \cos(2\alpha)) = 2(\frac{\pi}{12}) 	imes 2 \cos(\frac{\pi}{6}) = \frac{\pi}{6} 	imes 2 	imes \frac{\sqrt{3}}{2} = \frac{\pi}{2\sqrt{3}}$.

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

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