1. The maximum value of $y=2 x+\sqrt{6-x}$ is
$=.1 .12 \frac{1}{8}$.
Let $\sqrt{6-x}=t(t \geqslant 0)$, then $x=6-t^{2}$.
Thus $y=-2 t^{2}+t+12=-2\left(t-\frac{1}{4}\right)^{2}+12 \frac{1}{8}$.
Therefore, when $t=\frac{1}{4}$, i.e., $x=\frac{95}{16}$, $y_{\text {max }}=12 \frac{1}{8}$.