11.6. What is the minimum value of the ratio of the volumes of a cone and a cylinder circumscribed about the same sphere?
11.6. Let $r$ be the radius of the given sphere. If the axial section of the cone is an isosceles triangle with height $h$ and base $2a$, then $ah = S = r\left(a + \sqrt{h^2 + a^2}\right)$. Therefore, $a^2(h - r)^2 = r^2(h^2 + a^2)$, i.e., $a^2 = r^2 h^2 / (h - 2r)$. Hence, the volume of the cone is $\pi r^2 h^2 / 3(h - 2r)$. Since

$$
\frac{d}{dh}\left(\frac{h^2}{h - 2r}\right) = -\frac{4rh - h^2}{(h - 2r)^2}
$$

the volume of the cone is minimal when $h = 4r$. In this case, the ratio of the volumes of the cone and the cylinder is $4 / 3$.