3. Let $R$ be the triangular region (including the boundaries of the triangle) in the plane with vertices at $A(4,1), B(-1,-6), C(-3,2)$. Find the maximum and minimum values of the function $4x - 3y$ as $(x, y)$ varies over $R$. (You must prove your assertion)

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To find the maximum and minimum values of the function \( f(x, y) = 4x - 3y \) over the triangular region \( R \) with vertices \( A(4,1) \), \( B(-1,-6) \), and \( C(-3,2) \), we need to evaluate \( f(x, y) \) at the vertices and along the boundaries of the triangle.

### Step 1: Evaluate \( f(x, y) \) at the vertices
1. At \( A(4,1) \):
   \[
   f(4,1) = 4(4) - 3(1) = 16 - 3 = 13
   \]

2. At \( B(-1,-6) \):
   \[
   f(-1,-6) = 4(-1) - 3(-6) = -4 + 18 = 14
   \]

3. At \( C(-3,2) \):
   \[
   f(-3,2) = 4(-3) - 3(2) = -12 - 6 = -18
   \]

### Step 2: Evaluate \( f(x, y) \) along the boundaries
We need to check the function along the line segments \( AB \), \( BC \), and \( CA \).

#### Line segment \( AB \):
The equation of the line \( AB \) can be found using the points \( A(4,1) \) and \( B(-1,-6) \):
\[
\text{slope of } AB = \frac{-6 - 1}{-1 - 4} = \frac{-7}{-5} = \frac{7}{5}
\]
The equation of the line is:
\[
y - 1 = \frac{7}{5}(x - 4) \implies y = \frac{7}{5}x - \frac{28}{5} + 1 \implies y = \frac{7}{5}x - \frac{23}{5}
\]
Substitute \( y = \frac{7}{5}x - \frac{23}{5} \) into \( f(x, y) \):
\[
f(x, y) = 4x - 3\left(\frac{7}{5}x - \frac{23}{5}\right) = 4x - \frac{21}{5}x + \frac{69}{5} = \frac{20x - 21x + 69}{5} = \frac{-x + 69}{5}
\]
Evaluate \( f(x, y) \) at the endpoints \( A(4,1) \) and \( B(-1,-6) \):
\[
f(4,1) = 13 \quad \text{and} \quad f(-1,-6) = 14
\]

#### Line segment \( BC \):
The equation of the line \( BC \) can be found using the points \( B(-1,-6) \) and \( C(-3,2) \):
\[
\text{slope of } BC = \frac{2 - (-6)}{-3 - (-1)} = \frac{8}{-2} = -4
\]
The equation of the line is:
\[
y + 6 = -4(x + 1) \implies y = -4x - 4 - 6 \implies y = -4x - 10
\]
Substitute \( y = -4x - 10 \) into \( f(x, y) \):
\[
f(x, y) = 4x - 3(-4x - 10) = 4x + 12x + 30 = 16x + 30
\]
Evaluate \( f(x, y) \) at the endpoints \( B(-1,-6) \) and \( C(-3,2) \):
\[
f(-1,-6) = 14 \quad \text{and} \quad f(-3,2) = -18
\]

#### Line segment \( CA \):
The equation of the line \( CA \) can be found using the points \( C(-3,2) \) and \( A(4,1) \):
\[
\text{slope of } CA = \frac{1 - 2}{4 - (-3)} = \frac{-1}{7}
\]
The equation of the line is:
\[
y - 2 = -\frac{1}{7}(x + 3) \implies y = -\frac{1}{7}x - \frac{3}{7} + 2 \implies y = -\frac{1}{7}x + \frac{11}{7}
\]
Substitute \( y = -\frac{1}{7}x + \frac{11}{7} \) into \( f(x, y) \):
\[
f(x, y) = 4x - 3\left(-\frac{1}{7}x + \frac{11}{7}\right) = 4x + \frac{3}{7}x - \frac{33}{7} = \frac{28x + 3x - 33}{7} = \frac{31x - 33}{7}
\]
Evaluate \( f(x, y) \) at the endpoints \( C(-3,2) \) and \( A(4,1) \):
\[
f(-3,2) = -18 \quad \text{and} \quad f(4,1) = 13
\]

### Conclusion
The maximum value of \( f(x, y) \) over the region \( R \) is \( 14 \) and the minimum value is \( -18 \).

\[
\boxed{14 \text{ and } -18}
\]
Let $P\left(x_{0}, y_{0}\right)$ be any point on $R$, and draw the line $z_{P}=4 x-3 y$ through $P$. Then for any point $M(x, y)$ on this line, $z_{M}=z_{P}$. Since the line intersects the boundary of $R$, the maximum and minimum values of $z_{P}$ must be obtained at $A, B, C$. Given that $z_{A}=13, z_{B}=14, z_{C}=-18$, the maximum value of the function $4 x-3 y$ is 14, and the minimum value is -18.