We have to maximize Z = 3x + 4y
First, we will convert the given inequations into equations, we obtain the following equations:
2x + 2y = 80, 2x + 4y = 120
Region represented by 2x + 2y ≤ 80:
The line 2x + 2y = 80 meets the coordinate axes at A(40, 0)A40, 0 and B(0, 40)B0, 40 respectively. By joining these points we obtain the line 2x + 2y = 80.
Clearly (0,0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80.
Region represented by 2x + 4y ≤ 120:
The line 2x + 4y = 120 meets the coordinate axes at C(60, 0)C60, 0 and D(0, 30)D0, 30 respectively. By joining these points we obtain the line 2x + 4y ≤ 120.
Clearly (0,0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120.
The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:
The corner points of the feasible region are O(0, 0), A(40, 0)A40, 0, E(20, 20)E20, 20 and D(0, 30)D0, 30.
The values of Z at these corner points are as follows:
| Corner point | Z = 3x + 4y |
| O(0, 0) | 3 × 0 + 4 × 0 = 0 |
| A(40, 0) | 3× 40 + 4 × 0 = 120 |
| E(20, 20) | 3 × 20 + 4 × 20 = 140 |
| D(0, 30) | 10 × 0 + 4 ×30 = 120 |
We see that the maximum value of the objective function Z is 140 which is at E(20, 20)E20, 20 that means at x = 20 and y = 20.
Thus, the optimal value of Z is 140.
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