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(A) Let the height of the cone be $h$ and the radius of its base be $r$. The sphere has radius $R$. From the geometry of the cone inscribed in a spher

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$8: 27$

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(A) Let the height of the cone be $h$ and the radius of its base be $r$. The sphere has radius $R$. From the geometry of the cone inscribed in a sphere,we have the relation $r^2 = R^2 - (h - R)^2 = 2hR - h^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (2hR - h^2) h = \frac{1}{3} \pi (2Rh^2 - h^3)$.To find the maximum volume,we differentiate $V$ with respect to $h$ and set it to zero:$\frac{dV}{dh} = \frac{1}{3} \pi (4Rh - 3h^2) = 0$.Since $h 
eq 0$,we have $4R - 3h = 0$,which gives $h = \frac{4R}{3}$.Thus,$k = \frac{4}{3}$.The volume of the cone at $h = \frac{4R}{3}$ is $V_{cone} = \frac{1}{3} \pi (2R(\frac{4R}{3})^2 - (\frac{4R}{3})^3) = \frac{1}{3} \pi (\frac{32R^3}{9} - \frac{64R^3}{27}) = \frac{1}{3} \pi (\frac{96R^3 - 64R^3}{27}) = \frac{32}{81} \pi R^3$.The volume of the sphere is $V_{sphere} = \frac{4}{3} \pi R^3$.The ratio of the volume of the cone to the volume of the sphere is $\frac{\frac{32}{81} \pi R^3}{\frac{4}{3} \pi R^3} = \frac{32}{81} 	imes \frac{3}{4} = \frac{8}{27}$.

If the height of a cone of greatest volume that can be inscribed in a sphere of radius $R$ is $kR$,then the ratio of the volume of the cone to the volume of the sphere is

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