Four, a gas station needs to manufacture a cylindrical oil storage tank with a volume of $20 \pi \mathrm{m}^{3}$. It is known that the iron plate used for the bottom costs 40 yuan per square meter, and the iron plate used for the side costs 32 yuan per square meter. If manufacturing loss is not considered, what should the radius of the base and the height of the storage tank be to minimize the material cost for manufacturing the storage tank?
Let the base radius of a cylindrical oil tank be $r \mathrm{~m}$, the height be $h \mathrm{~m}$, and the cost be $y$ yuan.
Then $\pi r^{2} h=20 \pi, \therefore h=\frac{20 \pi}{\pi r^{2}}=\frac{20}{r^{2}}$.
$$
\begin{array}{l}
\therefore y=2 \pi r^{2} \cdot 40+2 \pi r h \cdot 32=80 \pi r^{2}+64 \pi r h \\
=80 \pi r^{2}+64 \pi r \cdot \frac{20}{r^{2}}=80 \pi r^{2}+\frac{1290 \pi}{r} \\
=80 \pi r^{2}+\frac{640 \pi}{r}+\frac{640 \pi}{r} \\
\geqslant 3 \sqrt[3]{80 \pi r^{2} \cdot \frac{640 \pi}{r} \cdot \frac{640 \pi}{r}} \\
=960 \pi(\text { yuan }) .
\end{array}
$$

When and only when $80 \pi r^{2}=\frac{640 \pi}{r}$, i.e., $r=2, h=5$, the equality holds.
Answer: When the base radius of the oil tank is $2 \mathrm{~m}$ and the height is $5 \mathrm{~m}$, the material cost is the lowest.