We know the total surface area of a right circular cone is $\pi m^{2}$ and its volume is $\frac{1}{3} \pi a^{3}$. Find its height and the radius of its base. When the volume is constant, when will the surface area be minimum? If the surface area is constant, when will the volume be maximum? In the latter case, calculate the angle formed by the slant height and the axis of the cone.
Given that the height of the cone is $h$, the slant height is $l$, and the radius of the base is $r$, according to the problem:

$$
\pi r(r+l)=\pi m^{2}
$$

and

$$
\frac{\pi}{3} r^{2} h=\frac{\pi}{3} a^{3}
$$

From these two equations, after substituting $l=\sqrt{h^{2}+r^{2}}$, we obtain:

$$
\begin{aligned}
& r=\frac{1}{2} \sqrt{\frac{m^{3} \pm \sqrt{m^{6}-8 a^{6}}}{m}} \\
& h=\frac{m\left(m^{3} \mp \sqrt{m^{6}-8 a^{6}}\right)}{2 a^{3}}
\end{aligned}
$$

For $r$ and $h$ to be real, it must hold that:

$$
m^{6}-8 a^{6} \geqq 0
$$

which means

$$
\frac{m}{a} \geqq \sqrt{2}
$$

This also provides the answer to the second and third questions. If the volume is constant, and thus $a$ is constant, the minimum value that $m$ can take is $m=a \sqrt{2}$. In this case, the surface area is minimized. However, for a constant surface area, the maximum volume is achieved when $a=\frac{m}{\sqrt{2}}$.

In the latter case, for the angle $\alpha$ formed by the slant height and the axis:

$$
\operatorname{tg} \alpha=\frac{r}{h}=\frac{\frac{a}{2} \sqrt{2}}{2 a}=\frac{1}{4} \sqrt{2}
$$

from which

$$
\alpha=19^{\circ} 28^{\prime} 15^{\prime \prime}
$$

(Lajos Kertész.)

The problem was also solved by Dénes A., Kornis Ö., Roth M., Szabó I., Szabó K.