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(C) Let the radius of the cylinder be $r$ and its height be $h$. The sphere has a radius $a$ (since diameter is $2a$).In the right-angled triangle for

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$\frac{2 a}{\sqrt{3}}$

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(C) Let the radius of the cylinder be $r$ and its height be $h$. The sphere has a radius $a$ (since diameter is $2a$).In the right-angled triangle formed by the radius of the sphere,the radius of the cylinder,and half the height of the cylinder,we have by Pythagoras theorem:$r^2 + (h/2)^2 = a^2$$r^2 = a^2 - \frac{h^2}{4}$The volume $V$ of the cylinder is given by $V = \pi r^2 h$.Substituting $r^2$ in the volume formula:$V = \pi (a^2 - \frac{h^2}{4}) h = \pi (a^2 h - \frac{h^3}{4})$To find the maximum volume,we differentiate $V$ with respect to $h$:$\frac{dV}{dh} = \pi (a^2 - \frac{3h^2}{4})$Setting $\frac{dV}{dh} = 0$ for critical points:$a^2 - \frac{3h^2}{4} = 0$$3h^2 = 4a^2$$h^2 = \frac{4a^2}{3}$$h = \frac{2a}{\sqrt{3}}$To verify it is a maximum,we check the second derivative:$\frac{d^2V}{dh^2} = \pi (0 - \frac{6h}{4}) = -\frac{3\pi h}{2}  0$.Since the second derivative is negative,the volume is maximum at $h = \frac{2a}{\sqrt{3}}$.

The height of a right circular cylinder of maximum volume inscribed in a sphere of diameter $2 a$ is

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