2. Given that $x$, $y$, $z$ are all non-negative numbers, and satisfy
$$
y+z-1=4-y-2 z=x \text{. }
$$

If $w=2 x^{2}-2 y+z$, then the minimum value of $w$ is ).
(A) -1
(B) $\frac{23}{9}$
(C) $-\frac{1}{2}$
(D) 0
2. C.

From the given, we have $y=3 x-2 \geqslant 0, z=-2 x+3 \geqslant 0$. Thus, $\frac{2}{3} \leqslant x \leqslant \frac{3}{2}$.
And $w=2 x^{2}-2 y+z$ $=2 x^{2}-8 x+7=2(x-2)^{2}-1$.
Therefore, when $x=\frac{3}{2}$, $w_{\min }=-\frac{1}{2}$.