A right circular cone is inscribed in a spher | vedclass

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(D) Let $R = 3$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.Let $x$ be the distance f

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$	an ^{-1}\left(\frac{1}{\sqrt{2}}ight)$

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(D) Let $R = 3$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.Let $x$ be the distance from the center of the sphere to the base of the cone. Then $h = R + x = 3 + x$.In the right triangle formed by the radius of the sphere, the radius of the cone base, and the distance $x$, we have $r^2 + x^2 = R^2 = 3^2 = 9$, so $r^2 = 9 - x^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (9 - x^2)(3 + x)$.$V = \frac{\pi}{3} (27 + 9x - 3x^2 - x^3)$.To maximize $V$, we find $\frac{dV}{dx} = \frac{\pi}{3} (9 - 6x - 3x^2) = 0$.$x^2 + 2x - 3 = 0 \Rightarrow (x + 3)(x - 1) = 0$.Since $x$ is a distance, $x = 1$ (as $x 
eq -3$).For $x = 1$, $r^2 = 9 - 1^2 = 8$, so $r = \sqrt{8} = 2\sqrt{2}$.The height $h = 3 + 1 = 4$.The semi-vertical angle $	heta$ satisfies $	an 	heta = \frac{r}{h} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$.Thus, $	heta = 	an ^{-1}\left(\frac{1}{\sqrt{2}}ight)$.

$A$ right circular cone is inscribed in a sphere of radius $3$ units. If the volume of the cone is maximum, then the semi-vertical angle of the cone is

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