Example 6 When the volume of a cylindrical metal beverage can is fixed, how should its height and base radius be selected to minimize the material used?
Let the height and base radius of the cylinder be $h$ and $R$, respectively. Then the surface area: $S=2 \pi R h + 2 \pi R^{2}$. From $V=\pi R^{2} h$, we get $h=\frac{V}{\pi R^{2}}$, so $S(R)=2 \pi R \cdot \frac{V}{\pi R^{2}} + 2 \pi R^{2} = \frac{2 V}{R} + 2 \pi R^{2}$. Let $S^{\prime}(R)=-\frac{2 V}{R^{2}} + 4 \pi R = 0$, we get $R=\sqrt[3]{\frac{V}{2 \pi}}$, thus $h=\frac{V}{\pi R^{2}} = 2 \sqrt[3]{\frac{V}{2 \pi}}$, i.e., $h=2 R$. Since $S(R)$ has a minimum but no maximum and has only one extremum, it is the minimum. Answer: When the height of the can is equal to the diameter of the base, the material used is minimized.