The cost of the iron plate used for the bottom is 40 yuan per square meter, and the cost of the iron plate used for the sides is 32 yuan per square meter. If we ignore the loss in production, what should the radius of the base and the height of the oil tank be to minimize the material cost of making the oil tank?
Let the radius of the base of a cylindrical oil tank be $r \mathrm{~m}$, the height be $h \mathrm{~m}$, and the cost be $y$ yuan. According to the problem, we have $\pi r^{2} h=20 \pi$, so $h=\frac{20}{r^{2}}$. Therefore,
$$
\begin{aligned}
y & =2 \pi r^{2} \cdot 40+2 \pi r h \cdot 32=80 \pi r^{2}+64 \pi r h=80 \pi r^{2}+64 \pi r \cdot \frac{20}{r^{2}} \\
& =80 \pi r^{2}+\frac{1280 \pi}{r}=80 \pi r^{2}+\frac{640 \pi}{r}+\frac{640 \pi}{r} \\
& \geqslant 3 \sqrt[3]{80 \pi r^{2} \cdot \frac{640 \pi}{r} \cdot \frac{640 \pi}{r}}=960 \pi(\text { yuan }) .
\end{aligned}
$$

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