(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 $g(m)$.
13 Let $t=\sqrt{100-m x}$, then $x=\frac{100-t^{2}}{m}$, so
$$
y=\frac{100-t^{2}}{m}+t=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} .
$$

When $t=\frac{m}{2}$, $y$ has the maximum value $\frac{100}{m}+\frac{m}{4}$, i.e., $g(m)=\frac{100}{m}+\frac{m}{4}$. Therefore,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt{\frac{100}{m} \times \frac{m}{4}}=10,
$$

the equality holds if and only if $m=20$, so when $m=20$, $g(m)$ has the minimum value 10.