In the Cartesian coordinate system $x 0 y$, there are points $A(3,3)$, $B(-2,1)$, $C(1,-2)$, and $T$ represents the set of all points inside and on the sides (including vertices) of $\triangle ABC$. Try to find the range of values for the bivariate function
$$
f(x, y)=\max \{2 x+y, \min \{3 x-y, 4 x+y\}\}
$$
(points $(x, y) \in T$).
As shown in Figure 5.
$l_{B C}: y=-x-1$,
$l_{A B}: y=\frac{2}{5} x+\frac{9}{5}$,
$l_{A C}: y=\frac{5}{2} x-\frac{9}{2}$.
We will discuss the following cases:
(1) If $4 x+y \geqslant 3 x-y$, then $x+2 y \geqslant 0$.

In this case, $f(x, y)=\max \{2 x+y, 3 x-y\}($ point $(x, y) \in$ $T$, and $x+2 y \geqslant 0$ ).
Let the intersection of the line $x+2 y=0$ and $A C$ be point $D\left(\frac{3}{2},-\frac{3}{4}\right)$.
If $2 x+y \geqslant 3 x-y$, then $y \geqslant \frac{x}{2}$, in this case, $f(x, y)=2 x+y$. Let the intersection of the line $y=\frac{x}{2}$ and $A D$ be point $E\left(\frac{9}{4}, \frac{9}{8}\right)$. Point $(x, y)$ is within quadrilateral $A B O E$ (including the boundary), it is easy to see that $f(x, y)=2 x+y$ reaches its maximum value at point $A$ and its minimum value at point $B$, so, $-3 \leqslant f(x, y) \leqslant 9$.

If $2 x+y \leqslant 3 x-y$, then $y \leqslant \frac{x}{2}$, in this case, $f(x, y)=3 x-y$. Point $(x, y)$ is within $\triangle E O D$ (including the boundary), it is easy to see that $f(x, y)=3 x-y$ reaches its maximum value at point $E$ and its minimum value at point $O$, so, $0 \leqslant f(x, y) \leqslant \frac{45}{8}$.
(2) If $4 x+y \leqslant 3 x-y$, then $x+2 y \leqslant 0$.
In this case, $f(x, y)=\max \{2 x+y, 4 x+y\}($ point $(x, y) \in$ $T$, and $x+2 y \leqslant 0$ ), point $(x, y)$ is within $\triangle B C D$ (including the boundary).

If $2 x+y \geqslant 4 x+y$, then $x \leqslant 0$. Let the intersection of the line $x=0$ and $B C$ be point $F(0,-1)$, in this case, point $(x, y)$ is within $\triangle B O F$ (including the boundary), $f(x, y)=2 x+y$. It is easy to see that $f(x, y)=2 x+y$ reaches its maximum value at point $O$ and its minimum value at point $B$, so, $-3 \leqslant f(x, y) \leqslant 0$.

If $2 x+y \leqslant 4 x+y$, then $x \geqslant 0$. Point $(x, y)$ is within quadrilateral OFCD (including the boundary), $f(x, y)=4 x+y$. It is easy to see that $f(x, y)=4 x+y$ reaches its maximum value at point $D$ and its minimum value at point $F$, so, $-1 \leqslant f(x, y) \leqslant \frac{21}{4}$.
In summary, the range of $f(x, y)$ is $[-3,9]$.