21. For each real number $x$, let $f(x)$ be the minimum of the numbers $4 x+1, x+2$ and $-2 x+4$. Determine the maximum value of $6 f(x)+2012$.
21. Answer: 2028
Let $L_{1}, L_{2}, L_{3}$ represent the three lines $y=4 x+1, y=x+2$ and $y=-2 x+4$ respectively.
Observe that $L_{1}$ and $L_{2}$ intersects at $\left(\frac{1}{3}, \frac{7}{3}\right), L_{1}$ and $L_{3}$ intersects at $\left(\frac{1}{2}, 3\right)$, and $L_{2}$ and $L_{3}$ intersects at $\left(\frac{2}{3}, \frac{8}{3}\right)$. Thus
$$
f(x)=\left\{\begin{array}{ll}
4 x+1, & x\frac{2}{3} .
\end{array}\right.
$$

Thus the maximum value of $f(x)$ is $\frac{8}{3}$ and the maximum value of $6 f(x)+2012$ is 2028.