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(B) Let $l$ be the slant height and $\alpha$ be the semi-vertical angle of the right circular cone. Let $h$ be the height and $r$ be the radius of the

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$tan^{-1}(\sqrt{2})$

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(B) Let $l$ be the slant height and $\alpha$ be the semi-vertical angle of the right circular cone. Let $h$ be the height and $r$ be the radius of the base.Then $h = l \cos \alpha$ and $r = l \sin \alpha$.The volume $V$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.Substituting the values of $r$ and $h$,we get $V = \frac{1}{3} \pi (l \sin \alpha)^2 (l \cos \alpha) = \frac{1}{3} \pi l^3 \sin^2 \alpha \cos \alpha$.To find the maximum volume,differentiate $V$ with respect to $\alpha$:$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [\sin^2 \alpha (-\sin \alpha) + \cos \alpha (2 \sin \alpha \cos \alpha)]$$\frac{dV}{d\alpha} = \frac{1}{3} \pi l^3 [-\sin^3 \alpha + 2 \sin \alpha \cos^2 \alpha]$.Setting $\frac{dV}{d\alpha} = 0$ for critical points:$-\sin^3 \alpha + 2 \sin \alpha (1 - \sin^2 \alpha) = 0$$\sin \alpha (2 - 3 \sin^2 \alpha) = 0$.Since $\alpha 
eq 0$,we have $2 - 3 \sin^2 \alpha = 0$,which implies $\sin^2 \alpha = \frac{2}{3}$.Using $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{2}{3} = \frac{1}{3}$,we find $	an^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{2/3}{1/3} = 2$.Thus,$	an \alpha = \sqrt{2}$,or $\alpha = 	an^{-1}(\sqrt{2})$.Checking the second derivative $\frac{d^2V}{d\alpha^2}$ confirms that the volume is maximum at this value.

Find the semi-vertical angle of a right circular cone of a given slant height,if the volume of the cone is maximum.

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