5. If the volume of a cylinder is numerically equal to its surface area, then the minimum possible value of the volume of the cylinder is
A. $48 \pi$
B. $50 \pi$
C. $54 \pi$
D. $66 \pi$
5. $\mathrm{C}$ Let the radius and height of the cylinder be $r, h$, respectively. According to the problem, we have $\pi r^{2} h=2 \pi r^{2}+2 \pi r h$, solving this gives
$$
h=\frac{2 r}{r-2} .
$$

Thus, $V=\pi r^{2} h=2 \pi \cdot \frac{r^{3}}{r-2}$. Let $r-2=x(x>0)$,
then $V=2 \pi \cdot \frac{(x+2)^{3}}{x}=2 \pi \cdot \frac{(1+1+x)^{3}}{x} \geqslant 2 \pi \cdot \frac{(3 \sqrt[3]{1 \cdot 1 \cdot x})^{3}}{x}=54 \pi$, with equality holding if and only if $x=1$, i.e., $r=3$. Therefore, $V_{\min }=54 \pi$.