Among the unit volume frustums of revolution, which one has the minimum surface area?
Solution. Let the radius of the base circle of the cone be $r$, its height be $m$, and its slant height be $a$. Then $V=\frac{r^{2} \pi m}{3}=1, r^{2}=\frac{3}{m \pi}$. According to the Pythagorean theorem,

$$
a=\sqrt{r^{2}+m^{2}}=\sqrt{\frac{3}{m \pi}+m^{2}}
$$

The surface area of the cone is:

$$
A=r^{2} \pi+r \pi a=\frac{3}{m \pi} \cdot \pi+\sqrt{\frac{3}{m \pi}} \cdot \pi \cdot \sqrt{\frac{3}{m \pi}+m^{2}}=\frac{3}{m}+\pi \cdot \sqrt{\frac{9}{m^{2} \pi^{2}}+\frac{3 m}{\pi}}
$$

Rearranging:

$$
A-\frac{3}{m}=\sqrt{\frac{9}{m^{2}}+3 m \pi}
$$

Since $A \geq \frac{3}{m}$, we can square both sides:

$$
A^{2}-\frac{6 A}{m}+\frac{9}{m^{2}}=\frac{9}{m^{2}}+3 m \pi, \quad A^{2} m-6 A=3 m^{2} \pi, \quad 3 m^{2} \pi-A^{2} m+6 A=0
$$

Thus,

$$
m_{1,2}=\frac{A^{2} \pm \sqrt{A^{4}-72 A \pi}}{6 \pi}
$$

The discriminant must be non-negative, so

$$
A^{4}-72 A \pi \geq 0, \quad \text { i.e., } \quad A \geq \sqrt[3]{72 \pi}
$$

Therefore, the minimum surface area is $A=\sqrt[3]{72 \pi}$.

In this case, $D=0$, so

$$
\begin{aligned}
& m=\frac{A^{2}}{6 \pi}=\frac{\sqrt[3]{(72 \pi)^{2}}}{6 \pi}=\sqrt[3]{\frac{72^{2} \cdot \pi^{2}}{6^{3} \cdot \pi^{3}}}=2 \cdot \sqrt[3]{\frac{3}{\pi}} \\
& r^{2}=\frac{3}{m \pi}=\frac{3}{\pi \cdot 2 \cdot \sqrt[3]{\frac{3}{\pi}}}=\sqrt[3]{\frac{3^{3}}{\pi^{3} \cdot 2^{3} \cdot \frac{3}{\pi}}}=\sqrt[3]{\frac{9}{2^{3} \cdot \pi^{2}}}=\frac{1}{2} \cdot \sqrt[3]{\frac{9}{\pi^{2}}}
\end{aligned}
$$

Thus, $r=\frac{1}{\sqrt{2}} \cdot \sqrt[3]{\frac{3}{\pi}}$

Therefore, among the cones with a unit volume, the one with the smallest surface area has a height of $2 \cdot \sqrt[3]{\frac{3}{\pi}}$ and a base radius of $\frac{1}{\sqrt{2}} \cdot \sqrt[3]{\frac{3}{\pi}}$.