The minimum value of $Z = 3x + y$, subject to the constraints $2x + 3y \leq 6, x + y \geq 1, x \geq 0, y \geq 0$ is....

The objective function is $Z = 3x + y$.

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

The feasible region is bounded by the lines $2x + 3y = 6$, $x + y = 1$, $x = 0$, and $y = 0$.

The corner points of the feasible region are the intersections of these lines with the coordinate axes:
For $x + y = 1$, the intercepts are $(1, 0)$ and $(0, 1)$.
For $2x + 3y = 6$, the intercepts are $(3, 0)$ and $(0, 2)$.

Thus, the vertices of the feasible region are $(1, 0)$, $(3, 0)$, $(0, 2)$, and $(0, 1)$.

Evaluating $Z = 3x + y$ at each corner point:
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 minimum value of $Z$ is $1$.