22. If $x, y \in \mathbf{R}$, and satisfy $\sqrt{x+2}+\sqrt{y-5}$ $=6$, then the minimum value of $x+2 y$ is $\qquad$ , and the maximum value is $\qquad$ _.
22.32,80.

Let $\sqrt{x+2}=a, \sqrt{y-5}=b$, then $a+b=6$.
Thus, $x+2 y=a^{2}-2+2\left(b^{2}+5\right)$
$$
=3(b-2)^{2}+32(0 \leqslant b \leqslant 6) \text {. }
$$

When $b=2$, $x+2 y$ has a minimum value of 32;
When $b=6$, $x+2 y$ has a maximum value of 80.