Minimize Z = 5x + y subject to 3x + y >= 12, x + y >= 8, x >= 0, y >= 0.

First, consider the boundary lines for the inequalities:

1. For $3x + y = 12$:

   • If $x = 0$, then $y = 12$. Point: $(0, 12)$
   • If $y = 0$, then $3x = 12 \Rightarrow x = 4$. Point: $(4, 0)$ Since $3x + y \ge 12$, the feasible region for this constraint is above or to the right of the line.

2. For $x + y = 8$:

   • If $x = 0$, then $y = 8$. Point: $(0, 8)$
   • If $y = 0$, then $x = 8$. Point: $(8, 0)$ Since $x + y \ge 8$, the feasible region for this constraint is above or to the right of the line.

3. The constraints $x \ge 0$ and $y \ge 0$ mean the feasible region is in the first quadrant.

The corner points are the intersections of the boundary lines.

• Intersection of $x = 0$ and $3x + y = 12$: Substitute $x = 0$ into $3x + y = 12 \Rightarrow 3(0) + y = 12 \Rightarrow y = 12$. Corner point: $(0, 12)$

• Intersection of $y = 0$ and $x + y = 8$: Substitute $y = 0$ into $x + y = 8 \Rightarrow x + 0 = 8 \Rightarrow x = 8$. Corner point: $(8, 0)$

• Intersection of $3x + y = 12$ and $x + y = 8$: Subtract the second equation from the first: $(3x + y) - (x + y) = 12 - 8$ $2x = 4$ $x = 2$ Substitute $x = 2$ into $x + y = 8$: $2 + y = 8$ $y = 6$ Corner point: $(2, 6)$

The corner points of the feasible region are $(0, 12)$, $(2, 6)$, and $(8, 0)$.

• At $(0, 12)$: $Z = 5(0) + 12 = 0 + 12 = 12$

• At $(2, 6)$: $Z = 5(2) + 6 = 10 + 6 = 16$

• At $(8, 0)$: $Z = 5(8) + 0 = 40 + 0 = 40$

Comparing the values of $Z$ at the corner points: $12, 16, 40$.
The minimum value is 12.

The minimum value of $Z$ is $\boxed{12}$.