Question

Solve the following linear programming problem graphically:
Maximize \[ Z = 8000x + 12000y \] Subject to the constraints
\[ 3x + 4y \le 60 \] \[ x + 3y \le 30 \] \[ x \ge 0,\; y \ge 0 \]

Hint:

In graphical linear programming problems, the optimal value of the objective function always occurs at one of the corner points (vertices) of the feasible region.

Answer 1

Solve the following linear programming problem graphically:
Maximize \[ Z = 8000x + 12000y \] Subject to the constraints
\[ 3x + 4y \le 60 \] \[ x + 3y \le 30 \] \[ x \ge 0,\; y \ge 0 \]

Step 1: Plot the constraints on the graph

We have the following constraints:

• For the first constraint \(3x + 4y \le 60\), rewrite the equation as \(y = \frac{60 - 3x}{4}\). This is a straight line with a y-intercept of 15 (when \(x = 0\)) and an x-intercept of 20 (when \(y = 0\)).
• For the second constraint \(x + 3y \le 30\), rewrite the equation as \(y = \frac{30 - x}{3}\). This is a straight line with a y-intercept of 10 (when \(x = 0\)) and an x-intercept of 30 (when \(y = 0\)).

Both lines should be drawn on the graph. The feasible region is the area that satisfies all the constraints, including the non-negativity conditions (\(x \ge 0\), \(y \ge 0\)).

Step 2: Find the corner points

The corner points of the feasible region can be found by solving the system of equations for the constraints:

1. Intersection of the lines \(3x + 4y = 60\) and \(x + 3y = 30\): \[ 3x + 4y = 60 \quad \text{(1)} \] \[ x + 3y = 30 \quad \text{(2)} \] Multiply equation (2) by 3: \[ 3x + 9y = 90 \quad \text{(3)} \] Subtract equation (1) from equation (3): \[ (3x + 9y) - (3x + 4y) = 90 - 60 \] \[ 5y = 30 \quad \Rightarrow \quad y = 6. \] Substitute \(y = 6\) into equation (2): \[ x + 3(6) = 30 \quad \Rightarrow \quad x = 12. \] Therefore, the intersection point is \((12, 6)\). 2. The other corner points are the intercepts: \[ (0, 15) \quad \text{from} \quad 3x + 4y = 60. \] \[ (30, 0) \quad \text{from} \quad x + 3y = 30. \]

Step 3: Evaluate the objective function at each corner point

We now evaluate \(Z = 8000x + 12000y\) at each corner point: - At \((0, 15)\): \[ Z = 8000(0) + 12000(15) = 180000. \] - At \((30, 0)\): \[ Z = 8000(30) + 12000(0) = 240000. \] - At \((12, 6)\): \[ Z = 8000(12) + 12000(6) = 96000 + 72000 = 168000. \]

Step 4: Conclusion

The maximum value of \(Z\) occurs at the corner point \((30, 0)\), and the maximum value of the objective function is: \[ Z = 240000. \]