Solutions

The volume of a right circular cylinder can be expressed as

$$V = h\pi r^2$$

assuming it has radius $r$ and height $h$. Its total surface area is

$$\begin{align*} a &= 2\pi r^2 + 2\pi rh \\ &= 2\pi r^2 + 2\pi r \left(\frac{V}{\pi r^2}\right) \\ &= 2\pi r^2 + \frac{2V}{r}. \end{align*}$$

Consider the interval of possible radii $r \in (0, \infty)$. Since

$$\lim_{r \to 0} a = \lim_{r \to \infty} a = \infty$$

and the function $a$ is continuous over $(0, \infty)$, we know that there is a minimum over $(0, \infty)$. The critical point is

$$\begin{align*} \frac{da}{dr} = 0 &\implies 4\pi r - \frac{2V}{r^2} = 0 \\ &\implies r^3 = \frac{V}{2\pi} \\ &\implies r = \sqrt[3]{\frac{V}{2\pi}}. \end{align*}$$

And it corresponds to the minimum by the interior extremum theorem since it is the only admissible critical point.