The lateral edge of a regular rectangular pyr | vedclass

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(C) Let the height of the pyramid be $h$ and the distance from the center of the base to a vertex be $x$. Given the lateral edge $a$,we have $h = a \s

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$\cot^{-1}\sqrt{2}$

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(C) Let the height of the pyramid be $h$ and the distance from the center of the base to a vertex be $x$. Given the lateral edge $a$,we have $h = a \sin \alpha$ and $x = a \cos \alpha$.The base is a square with diagonal $2x$. The area of the square base is $A = \frac{1}{2} 	imes (	ext{diagonal})^2 = \frac{1}{2} 	imes (2x)^2 = 2x^2$.The volume $V$ of the pyramid is $V = \frac{1}{3} A h = \frac{1}{3} (2x^2) h$.Substituting $x = a \cos \alpha$ and $h = a \sin \alpha$:$V(\alpha) = \frac{2}{3} (a \cos \alpha)^2 (a \sin \alpha) = \frac{2}{3} a^3 \sin \alpha \cos^2 \alpha$.To maximize $V$,we differentiate with respect to $\alpha$ and set to $0$:$V'(\alpha) = \frac{2}{3} a^3 [\cos \alpha \cdot \cos^2 \alpha + \sin \alpha \cdot 2 \cos \alpha (-\sin \alpha)] = 0$.$\cos^3 \alpha - 2 \sin^2 \alpha \cos \alpha = 0$.Since $\cos \alpha 
eq 0$,we have $\cos^2 \alpha = 2 \sin^2 \alpha$,which implies $	an^2 \alpha = \frac{1}{2}$,or $	an \alpha = \frac{1}{\sqrt{2}}$.Thus,$\alpha = 	an^{-1}(\frac{1}{\sqrt{2}}) = \cot^{-1}(\sqrt{2})$.

The lateral edge of a regular rectangular pyramid is $a \text{ cm}$ long. The lateral edge makes an angle $\alpha$ with the plane of the base. The value of $\alpha$ for which the volume of the pyramid is greatest,is

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