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(A) Let the radius of the sphere be $R$,so the diameter is $2R$.Let the height of the cone be $y$ and the radius of the base of the cone be $x$.From t

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$2/3$

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(A) Let the radius of the sphere be $R$,so the diameter is $2R$.Let the height of the cone be $y$ 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 $|y - R|$.Using the Pythagorean theorem in the triangle formed by the radius of the sphere,the radius of the cone,and the distance from the center: $x^2 + (y - R)^2 = R^2$.$x^2 = R^2 - (y - R)^2 = R^2 - (y^2 - 2Ry + R^2) = 2Ry - y^2$.The volume of the cone is $V = \frac{1}{3}\pi x^2 y = \frac{1}{3}\pi (2Ry - y^2)y = \frac{1}{3}\pi (2Ry^2 - y^3)$.To find the maximum volume,differentiate $V$ with respect to $y$: $\frac{dV}{dy} = \frac{1}{3}\pi (4Ry - 3y^2)$.Setting $\frac{dV}{dy} = 0$,we get $y(4R - 3y) = 0$. Since $y 
eq 0$,we have $y = \frac{4}{3}R$.The second derivative $\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 6y)$. At $y = \frac{4}{3}R$,$\frac{d^2V}{dy^2} = \frac{1}{3}\pi (4R - 8R) = -\frac{4}{3}\pi R The ratio of the height of the cone $(y)$ to the diameter of the sphere $(2R)$ is $\frac{y}{2R} = \frac{4/3 R}{2R} = \frac{2}{3}$.

$A$ cone of maximum volume is inscribed in a given sphere. Then the ratio of the height of the cone to the diameter of the sphere is:

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