The maximum volume (in cubic units) of the cy | vedclass

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(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R = 12$. From the geometry of the sphere, we have the

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$768 \sqrt{3} \pi$

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(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R = 12$. From the geometry of the sphere, we have the relation $R^2 = r^2 + (h/2)^2$, which gives $12^2 = r^2 + \frac{h^2}{4}$. Thus, $r^2 = 144 - \frac{h^2}{4}$. The volume $V$ of the cylinder is given by $V = \pi r^2 h$. Substituting $r^2$, we get $V = \pi (144 - \frac{h^2}{4}) h = 144 \pi h - \frac{\pi}{4} h^3$. To find the maximum volume, we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = 144 \pi - \frac{3 \pi}{4} h^2$. Setting $\frac{dV}{dh} = 0$, we get $144 \pi = \frac{3 \pi}{4} h^2$, which implies $h^2 = 144 	imes \frac{4}{3} = 192$. So, $h = \sqrt{192} = 8 \sqrt{3}$. Now, $r^2 = 144 - \frac{192}{4} = 144 - 48 = 96$. The maximum volume is $V = \pi r^2 h = \pi 	imes 96 	imes 8 \sqrt{3} = 768 \sqrt{3} \pi$ cubic units.

The maximum volume (in cubic units) of the cylinder which can be inscribed in a sphere of radius $12$ units is

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