$4 \cdot 46$ If $x+y+z=30, 3x+y-z=50$, where $x, y, z$ are all non-negative numbers. Then the range of $M=5x+4y+2z$ is
(A) $100 \leqslant M \leqslant 110$.
(B) $110 \leqslant M \leqslant 120$.
(C) $120 \leqslant M \leqslant 130$.
(D) $130 \leqslant M \leqslant 140$.
(China Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1989)
[Solution]
$$
\begin{aligned}
M & =5 x+4 y+2 z=(x+y+z)+(4 x+3 y+z) \\
& =30+4 x+3 y+z,
\end{aligned}
$$

From the given equations, adding them yields $4 x+2 y=80$,
$$
\therefore \quad M=30+(4 x+2 y)+y+z=110+y+z,
$$

Subtracting the given equations yields $2 x-2 z=20$,
$$
\because \quad z \geqslant 0, \quad \therefore \quad x \geqslant 10.
$$

From the assumption $30-(y+z)=x \geqslant 10$, we have $y+z \leqslant 20$,
$$
\therefore \quad M=110+y+z \leqslant 130.
$$

From $4 x+2 y=80$ we have $y=40-2 x \geqslant 0$, so $x \leqslant 20$, thus $y+z \geqslant 10$, and $M=110+y+z \geqslant 120$. Therefore, the answer is $(C)$.