3. Determine the maximum value of the ratio of the volumes of the sphere and the cone circumscribed around it.

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Solution. Let $r$ denote the radius of the sphere, $R$ and $H$ the radius of the base and the height of the cone, respectively, $\alpha$ the angle of the slant height of the cone with the plane of the base, and $t=\operatorname{tg} \frac{\alpha}{2}$, as shown in the diagram on the right. Then,

$$
H=R \operatorname{tg} \alpha=\frac{2 R t}{1-t^{2}}, \quad r=R \operatorname{tg} \frac{\alpha}{2}=R t
$$

Therefore, the ratio of the volume $V$ of the sphere to the volume $V^{\prime}$ of the cone is

![](https://cdn.mathpix.com/cropped/2024_06_05_72b79cc38cc70f646757g-05.jpg?height=408&width=372&top_left_y=1632&top_left_x=1107)

$$
V: V^{\prime}=\frac{4}{3} \pi r^{3}: \frac{1}{3} \pi R^{2} H=2 t^{2}(1-t)^{2} \leq 2 \cdot \frac{1}{4}=\frac{1}{2}
$$

where the equality holds if and only if $t^{2}=\frac{1}{2}$.

Finally, the desired maximum value of the ratio of the volumes is $\frac{1}{2}$, and it is achieved when the slant angle is $\alpha=2 \operatorname{arctg} \frac{1}{\sqrt{2}}$.