Given $x, y \in \mathbf{R}$, and satisfy $\sqrt{x+2}+\sqrt{y-5}$ $=6$. Find the minimum and maximum values of $x+2y$.

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Solution 1: Since $\frac{\sqrt{x+2}}{6}+\frac{\sqrt{y-5}}{6}=1$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
\sqrt{x+2}=6 \cos ^{2} \alpha, \\
\sqrt{y-5}=6 \sin ^{2} \alpha .
\end{array}\right. \\
\text { Solving, we get }\left\{\begin{array}{l}
x=36 \cos ^{4} \alpha-2, \\
y=36 \sin ^{4} \alpha+5 .
\end{array}\right. \\
\text { Therefore, } x+2 y=36 \cos ^{4} \alpha-2+72 \sin ^{4} \alpha+10 \\
=36\left(1-\sin ^{2} \alpha\right)^{2}+72 \sin ^{4} \alpha+8 \\
=36\left[\left(\sqrt{3} \sin ^{2} \alpha-\frac{1}{\sqrt{3}}\right)^{2}+\frac{2}{3}\right]+8 .
\end{array}
$$

Thus, when $\sqrt{3} \sin ^{2} \alpha=\frac{1}{\sqrt{3}}$, i.e., $\sin ^{2} \alpha=\frac{1}{3}$, $x+2 y$ has a minimum value of 32; when $\sin ^{2} \alpha=1$, $x+2 y$ has a maximum value of 80.

Solution 2: Let $\sqrt{x+2}-\sqrt{y-5}=n, x+2 y$ $=m$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
\sqrt{x+2}+\sqrt{y-5}=6, \\
\sqrt{x+2}-\sqrt{y-5}=n
\end{array}\right. \\
\left\{\begin{array}{l}
\sqrt{x+2}=\frac{6+n}{2}, \\
\sqrt{y-5}=\frac{6-n}{2}, \\
-6 \leqslant n \leqslant 6 .
\end{array}\right.
\end{array}
$$

Solving, we get $x=\frac{(6+n)^{2}}{4}-2, y=\frac{(6-n)^{2}}{4}+5$.
Therefore, $m=\frac{(6+n)^{2}}{4}-2+\frac{(6-n)^{2}}{2}+10$, i.e., $3 n^{2}-12 n+140-4 m=0$.

Thus, the problem is converted to finding the range of $m$ such that the equation (1) has a solution in the interval $[-6,6]$.
Since the axis of symmetry is $n=2$, based on the graph, we have
$\left\{\begin{array}{l}f(2) \leqslant 0, \\ f(-6) \geqslant 0,\end{array}\right.$ Solving, we get $32 \leqslant m \leqslant 80$.
Therefore, the minimum value of $x+2 y$ is 32, and the maximum value is
80.

Solution 3: Let $s=\sqrt{x+2}, t=\sqrt{y-5}, x+$ $2 y=m$. Then
$$
\begin{array}{l}
x=s^{2}-2, y=t^{2}+5, \\
m=x+2 y=s^{2}+2 t^{2}+8 .
\end{array}
$$

Thus, the problem is converted to finding the range of $m$ such that the ellipse $\frac{s^{2}}{m-8}+\frac{t^{2}}{\frac{m-8}{2}}=1$ intersects the line $s+t=6$ in the first quadrant (including endpoints).

Since the slope of the line $s+t=6$ is -1, the extreme values of $m$ are at the points
$$
A(\sqrt{m-8}, 0) 、 B\left(0, \sqrt{\frac{m-8}{2}}\right)
$$

and where the line is tangent to the ellipse.
Substituting points $A$ and $B$ into $s+t=6$, we get
$m=44$ and $m=80$.
Substituting $s=6-t$ into the ellipse equation and simplifying, we get
$3 t^{2}-12 t-m+44=0$.
Since the line is tangent to the ellipse, $\Delta=0$, giving $m=32$ (when and only when $t=2, s=4$).

Therefore, the minimum value of $x+2 y$ is 32, and the maximum value is 80.

Solution 4: Since the endpoints of the curve $\sqrt{x+2}+\sqrt{y-5}=6$ are $C(-2,41)$ and $D(34,5)$, substituting them into $m=$ $x+2 y$, we get $m=80$ or $m=44$.

The slope of the tangent line $x+2 y=m$ is $-\frac{1}{2}$. From $\sqrt{y-5}=6-\sqrt{x+2}$, we get
$$
\begin{array}{l}
y=x-12 \sqrt{x+2}+43, \\
y^{\prime}=1-\frac{6}{\sqrt{x+2}}=-\frac{1}{2} .
\end{array}
$$

Solving, we get $x=14$.
Substituting $x=14$ into $\sqrt{x+2}+\sqrt{y-5}=6$, we get

the coordinates of the tangent point are $(14,9)$.
Therefore, $m=x+2 y=32$.
In summary, the maximum value of $x+2 y$ is 80, and the minimum value is