For each real number $m$, the function $f(x)=x^{2}-2 m x+8 m+4$ has a minimum value. What is the maximum of these minimum values?
A quadratic function of the form $f(x)=a x^{2}+b x+c$ with $a>0$ attains its minimum when $x=-\frac{b}{2 a}$.

One way to see this is to complete the square to get

$$
f(x)=a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}
$$

and note that since $a\left(x+\frac{b}{2 a}\right)^{2}$ is nonnegative, the minimum will occur when this quantity equals 0 .

In a similar way, if $a<0$, the function attains its maximum when $x=-\frac{b}{2 a}$.

For $f(x)=x^{2}-2 m x+8 m+4$, the minimum occurs when $x=-\frac{-2 m}{2}=m$. The minimum value of $f(x)$ is therefore

$$
f(m)=m^{2}-2 m(m)+8 m+4=-m^{2}+8 m+4
$$

This is a quadratic function in the variable $m$ with a negative leading coefficient.

Therefore, the maximum occurs when $m=-\frac{8}{2(-1)}=4$.

The maximum of these minimum values is $-4^{2}+8(4)+4=20$.

ANSWER: 20