13. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$

has a maximum value of $g(m)$, find the minimum value of $g(m)$.
Three, 13. Let $t=\sqrt{100-m x}$. Then $x=\frac{100-t^{2}}{m}$.
Hence $y=\frac{100-t^{2}}{m}+t$
$$
=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} \text {. }
$$

When $t=\frac{m}{2}$, $y$ has a maximum value $\frac{100}{m}+\frac{m}{4}$, that is,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt{\frac{100}{m} \times \frac{m}{4}}=10 .
$$

Therefore, when and only when $m=20$, $g(m)$ has a minimum value of 10.