357. Find the dimensions of a cylindrical closed tank with a given volume $v$ and the smallest total surface area.
Solution. Denoting the radius and height of the cylinder by $r$ and $h$, and its total surface area by $s$, we get

$$
s=2 \pi r h + 2 \pi r^{2}
$$

Here, the variables $r$ and $h$ are not independent but are related by the equation $v=\pi r^{2} h$, since according to the condition, the cylinder must have a given volume $v$. Solving this equation for $h$ and substituting it into the expression for the total surface area, we get

$$
s=2\left(\pi r^{2}+\frac{v}{r}\right)
$$

where $r$ varies in the interval $0 < r < +\infty$.
From this, it follows that the critical point $r=\sqrt[3]{\frac{v}{2 \pi}}$ is a point of minimum.

The function $s(r)$ is continuous in the interval $(0 ;+\infty)$. Therefore, according to property 1 of continuous functions, the unique minimum of the function $s$ in the interval $(0 ;+\infty)$ coincides with its smallest value in this interval.

When $r=\sqrt[3]{\frac{v}{2 \pi}}$, we get $h=\frac{v}{\pi r^{2}}=2 \sqrt[3]{\frac{v}{2 \pi}}=2 r$.

Therefore, a cylindrical closed container with any given volume will have the smallest total surface area when its axial section is a square.