The difference between the maximum value and the minimum value of the objective function $z = 3x + y$ subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, $y \geq 0$ is....

The given objective function is $z = 3x + y$.

The constraints are $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, and $y \geq 0$.

The feasible region is bounded by the lines $2x + 3y = 6$, $x + y = 1$, $x = 0$, and $y = 0$. The corner points of this feasible region can be found by finding the intersections of these lines with the coordinate axes.

For $2x + 3y = 6$, the intercepts are $(3, 0)$ and $(0, 2)$.
For $x + y = 1$, the intercepts are $(1, 0)$ and $(0, 1)$.

The corner points of the bounded feasible region are $(1, 0)$, $(3, 0)$, $(0, 2)$, and $(0, 1)$.

Evaluating the objective function $z = 3x + y$ at these corner points:
At $(1, 0)$, $z = 3(1) + 0 = 3$
At $(3, 0)$, $z = 3(3) + 0 = 9$
At $(0, 2)$, $z = 3(0) + 2 = 2$
At $(0, 1)$, $z = 3(0) + 1 = 1$

The maximum value of $z$ is $9$ and the minimum value is $1$.

The difference between the maximum and minimum values is $9 - 1 = 8$.

Answer: $8$