The volume of the largest possible right circ | vedclass

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(C) Let the radius of the sphere be $R = \sqrt{3}$. Let the height of the cylinder be $h$ and its radius be $r$.In the right-angled triangle formed by

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$4\pi$

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(C) Let the radius of the sphere be $R = \sqrt{3}$. Let the height of the cylinder be $h$ and its radius be $r$.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:$R^2 = r^2 + (h/2)^2$$(\sqrt{3})^2 = r^2 + \frac{h^2}{4}$$3 = r^2 + \frac{h^2}{4} \Rightarrow r^2 = 3 - \frac{h^2}{4}$The volume of the cylinder is $V = \pi r^2 h = \pi (3 - \frac{h^2}{4})h = 3\pi h - \frac{\pi h^3}{4}$.To maximize the volume,we find the derivative with respect to $h$ and set it to zero:$\frac{dV}{dh} = 3\pi - \frac{3\pi h^2}{4} = 0$$3\pi = \frac{3\pi h^2}{4} \Rightarrow h^2 = 4 \Rightarrow h = 2$.Substituting $h = 2$ into the volume formula:$V = \pi (3 - \frac{2^2}{4})(2) = \pi (3 - 1)(2) = 4\pi$.

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $R = \sqrt{3}$ is

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