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(A) Let the radius of the sphere be $R$. Let the height of the cone be $h$ and the radius of the base of the cone be $x$. From the geometry of the sph

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$2/3$

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(A) Let the radius of the sphere be $R$. Let the height of the cone be $h$ and the radius of the base of the cone be $x$. From the geometry of the sphere,the distance from the center of the sphere to the base of the cone is $|h - R|$. Using the Pythagorean theorem in the triangle formed by the sphere's radius,the cone's base radius,and the distance from the center,we have $x^2 + (h - R)^2 = R^2$. $x^2 = R^2 - (h^2 - 2hR + R^2) = 2hR - h^2$. The volume of the cone is $V = \frac{1}{3}\pi x^2 h = \frac{1}{3}\pi (2hR - h^2)h = \frac{1}{3}\pi (2Rh^2 - h^3)$. To maximize volume,differentiate with respect to $h$: $\frac{dV}{dh} = \frac{1}{3}\pi (4Rh - 3h^2)$. Setting $\frac{dV}{dh} = 0$,we get $h(4R - 3h) = 0$. Since $h 
eq 0$,$h = \frac{4}{3}R$. The diameter of the sphere is $D = 2R$. The ratio of the height of the cone to the diameter of the sphere is $\frac{h}{D} = \frac{\frac{4}{3}R}{2R} = \frac{4}{6} = \frac{2}{3}$.

$A$ cone of maximum volume is inscribed in a sphere of radius $R$. The ratio of the height of the cone to the diameter of the sphere is:

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