Seven, for a cylindrical can with a fixed volume, when the relationship between the diameter of the base and the height is what, the least amount of iron sheet is used?



When the volume of a cylindrical can is fixed, the relationship between the diameter of the base and the height that results in the least amount of iron sheet being used is what?
Seven, Solution Let the base radius be $\mathrm{r}$, height be $\mathrm{h}$, total surface area be $\mathrm{S}$, and volume be $\mathbf{V}$.
$$
\begin{array}{l}
\because V=\pi r^{2} h, \quad \therefore h:=\frac{V}{\pi r^{2}} . \\
S=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+\frac{2 V}{r}, \\
S^{\prime}=\frac{4 \pi r^{2}-2 V}{r^{2}} .
\end{array}
$$

Let $S^{\prime}=0$, we get the critical point $r=\sqrt[3]{\frac{\mathrm{V}}{2 \pi}}$.
$$
r=\sqrt[3]{\frac{V}{2 \pi}} \text { when } S^{\prime}>0,
$$
$\therefore$ when $\mathrm{r}=\sqrt[3]{\frac{V}{2 \pi}}$ then $S$ is minimized, at this time $\mathrm{h}=2 \mathrm{r}$.