Step-by-Step Solution

1. Graph the constraints:

   • $x + 2y \le 8$ can be graphed as $x + 2y = 8$. The intercepts are (8,0) and (0,4).
   • $3x + 2y \le 12$ can be graphed as $3x + 2y = 12$. The intercepts are (4,0) and (0,6).
   • $x \ge 0$ and $y \ge 0$ restrict the solution to the first quadrant.

2. Identify the feasible region:

   The feasible region is the area bounded by the lines $x + 2y = 8$, $3x + 2y = 12$, $x = 0$, and $y = 0$.

3. Find the corner points of the feasible region:

   • Intersection of $x + 2y = 8$ and $3x + 2y = 12$: Subtracting the first equation from the second gives $2x = 4$, so $x = 2$. Substituting $x = 2$ into $x + 2y = 8$ gives $2 + 2y = 8$, so $2y = 6$ and $y = 3$. The intersection point is (2, 3).
   • Other corner points are (0, 0), (4, 0), and (0, 4).

   The corner points are (0, 0), (4, 0), (0, 4), and (2, 3).

4. Evaluate the objective function at each corner point:

   • At (0, 0): $z = -3(0) + 4(0) = 0$
   • At (4, 0): $z = -3(4) + 4(0) = -12$
   • At (0, 4): $z = -3(0) + 4(4) = 16$
   • At (2, 3): $z = -3(2) + 4(3) = -6 + 12 = 6$

5. Determine the minimum value:

   The minimum value of $z$ is -12, which occurs at the point (4, 0).