Step 1: Write the given constraints.
The constraints of the linear programming problem are
\[
3x + 4y \le 60
\]
\[
x + 3y \le 30
\]
\[
x \ge 0, \quad y \ge 0
\]
Since \(x\) and \(y\) are non–negative, the feasible region will lie in the first quadrant.
Step 2: Convert inequalities into equations.
To draw the boundary lines, convert the inequalities into equations:
\[
3x + 4y = 60
\]
\[
x + 3y = 30
\]
Step 3: Find intercepts of the first line.
For \(3x + 4y = 60\):
If \(x = 0\),
\[
4y = 60
\]
\[
y = 15
\]
If \(y = 0\),
\[
3x = 60
\]
\[
x = 20
\]
Thus the line passes through points
\[
(0,15) \quad \text{and} \quad (20,0)
\]
Step 4: Find intercepts of the second line.
For \(x + 3y = 30\):
If \(x = 0\),
\[
3y = 30
\]
\[
y = 10
\]
If \(y = 0\),
\[
x = 30
\]
Thus the line passes through
\[
(0,10) \quad \text{and} \quad (30,0)
\]
Step 5: Determine the feasible region.
The feasible region is obtained by satisfying all the inequalities simultaneously in the first quadrant.
The corner points of the feasible region are
\[
(0,0), \quad (0,10), \quad (12,6), \quad (20,0)
\]
Step 6: Find intersection of the two lines.
Solve
\[
3x + 4y = 60
\]
\[
x + 3y = 30
\]
Multiply the second equation by 3:
\[
3x + 9y = 90
\]
Subtract the first equation:
\[
5y = 30
\]
\[
y = 6
\]
Substitute into \(x + 3y = 30\):
\[
x + 18 = 30
\]
\[
x = 12
\]
Thus intersection point is
\[
(12,6)
\]
Step 7: Compute the value of $Z$ at each corner point.
\[
Z = 8000x + 12000y
\]
At \((0,0)\):
\[
Z = 0
\]
At \((0,10)\):
\[
Z = 12000(10) = 120000
\]
At \((12,6)\):
\[
Z = 8000(12) + 12000(6)
\]
\[
Z = 96000 + 72000
\]
\[
Z = 168000
\]
At \((20,0)\):
\[
Z = 8000(20) = 160000
\]
Step 8: Determine the maximum value.
The maximum value of \(Z\) occurs at
\[
(12,6)
\]
Step 9: Final conclusion.
The optimal solution of the linear programming problem is
\[
x = 12, \quad y = 6
\]
and the maximum value of the objective function is
\[
Z_{\max} = 168000
\]
Final Answer:
\[
\boxed{x=12,\; y=6,\; Z_{\max}=168000}
\]