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(C) Let the radius of the sphere be $R=3$. Let the height of the cylinder be $h$ and its radius be $r$.From the geometry of the sphere and cylinder,we

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$ 2\sqrt{3} $

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(C) Let the radius of the sphere be $R=3$. Let the height of the cylinder be $h$ and its radius be $r$.From the geometry of the sphere and cylinder,we have $r^2 + (h/2)^2 = R^2 = 3^2 = 9$.Thus,$r^2 = 9 - \frac{h^2}{4}$.The volume of the cylinder is $V = \pi r^2 h = \pi (9 - \frac{h^2}{4}) h = \pi (9h - \frac{h^3}{4})$.To maximize the volume,we find the derivative with respect to $h$ and set it to zero:$\frac{dV}{dh} = \pi (9 - \frac{3h^2}{4}) = 0$.$9 = \frac{3h^2}{4} \Rightarrow h^2 = 12 \Rightarrow h = \sqrt{12} = 2\sqrt{3}$.Thus,the height of the cylinder of maximum volume is $2\sqrt{3}$.

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius $3$ is

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