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(B) Let $R$ be the radius of the sphere and $r$ and $h$ be the radius and height of the cylinder, respectively.From the geometry of the sphere and ins

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$2 \sqrt{3}$

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(B) Let $R$ be the radius of the sphere and $r$ and $h$ be the radius and height of the cylinder, respectively.From the geometry of the sphere and inscribed cylinder, we have the relation: $R^2 = r^2 + (h/2)^2$.Thus, $r^2 = R^2 - h^2/4$.The volume $V$ of the cylinder is given by $V = \pi r^2 h = \pi (R^2 - h^2/4) h = \pi R^2 h - \pi h^3/4$.To find the maximum volume, we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4}$.Setting $\frac{dV}{dh} = 0$, we get $\pi R^2 = \frac{3\pi h^2}{4}$, which implies $h^2 = \frac{4R^2}{3}$, so $h = \frac{2R}{\sqrt{3}}$.Given $R = 3 \ cm$, we have $h = \frac{2 	imes 3}{\sqrt{3}} = 2 \sqrt{3} \ cm$.Since $\frac{d^2V}{dh^2} = -\frac{6\pi h}{4}  0$, the volume is maximum at $h = 2 \sqrt{3} \ cm$.

The height (in $cm$) of a cylinder of the greatest volume that can be inscribed in a sphere of radius $3 \ cm$ is

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