A rectangle ABCD is inscribed in the region b | vedclass

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(B) The rectangle is symmetric about the line $x = \frac{\pi}{2}$.Let the distance from the center $x = \frac{\pi}{2}$ to the sides be $\alpha$.The $x

Vedclass · Accepted Answer

$\cot \alpha = \alpha$

Vedclass · Answer

(B) The rectangle is symmetric about the line $x = \frac{\pi}{2}$.Let the distance from the center $x = \frac{\pi}{2}$ to the sides be $\alpha$.The $x-$coordinates of the vertices $B$ and $C$ are $\frac{\pi}{2} - \alpha$ and $\frac{\pi}{2} + \alpha$ respectively.The height of the rectangle is $y = \sin\left(\frac{\pi}{2} + \alphaight) = \cos \alpha$.The width of the rectangle is $(\frac{\pi}{2} + \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha$.Thus,the area $A$ of the rectangle is $A = (2\alpha)(\cos \alpha) = 2\alpha \cos \alpha$.To find the maximum area,we differentiate $A$ with respect to $\alpha$:$\frac{dA}{d\alpha} = 2(\cos \alpha - \alpha \sin \alpha)$.Setting $\frac{dA}{d\alpha} = 0$ gives $\cos \alpha = \alpha \sin \alpha$,which simplifies to $\cot \alpha = \alpha$ (since $\alpha 
eq 0$ and $\sin \alpha 
eq 0$ in the interval $(0, \frac{\pi}{2})$).

$A$ rectangle $ABCD$ is inscribed in the region bounded by $y = \sin x$ and the $x-$axis for $x \in [0, \pi]$ (as shown in the figure). The area of the rectangle is maximum when $'\alpha'$ satisfies:

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