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(A) Let $h$ be the height of the cone inscribed in a sphere of radius $R$. Let the center of the sphere be $O$. The distance from the center $O$ to th

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$2/3$

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(A) Let $h$ be the height of the cone inscribed in a sphere of radius $R$. Let the center of the sphere be $O$. The distance from the center $O$ to the base of the cone is $|h-R|$.By the Pythagorean theorem,the radius $r$ of the base of the cone satisfies $r^2 + (h-R)^2 = R^2$,so $r^2 = R^2 - (h^2 - 2hR + R^2) = 2hR - h^2$.The volume $V$ of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{1}{3} \pi (2h^2R - h^3)$.To maximize $V$,we find the derivative with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (4hR - 3h^2)$.Setting $\frac{dV}{dh} = 0$ gives $h(4R - 3h) = 0$. Since $h 
eq 0$,we have $h = \frac{4R}{3}$.The diameter of the sphere is $D = 2R$.The ratio of the altitude of the cone to the diameter of the sphere is $\frac{h}{D} = \frac{4R/3}{2R} = \frac{4}{6} = \frac{2}{3}$.

The ratio of the altitude of the cone of greatest volume which can be inscribed in a given sphere to the diameter of the sphere is:

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