Question: MAX 3 x 1 + 4 x 2 ($ Profit) s.t. x 1 + 3 x 2 12 2 x 1 + x 2 8
| MAX | 3x1 + 4x2 ($ Profit) |
| s.t. | x1 + 3x2 12 |
| 2x1 + x 2 8 | |
| x1 3 | |
| x1, x2 0 | |
The Management Scientist provided the following solution output: OPTIMAL SOLUTION Objective Function Value = 20.000
| Variable | Value | Reduced Cost |
| X1 | 2.400 | 0.000 |
| X2 | 3.200 | 0.000 |
| Constraint | Slack/Surplus | Dual Price |
| 1 | 0.000 | 1.000 |
| 2 | 0.000 | 1.000 |
| 3 | 0.600 | 0.000 |
OBJECTIVE COEFFICIENT RANGES
| Variable | Lower Limit | Current Value | Upper Limit |
| X1 | 1.333 | 3.000 | 8.000 |
| X2 | 1.500 | 4.000 | 9.000 |
RIGHT HAND SIDE RANGES
| Constraint | Lower Limit | Current Value | Upper Limit |
| 1 | 9.000 | 12.000 | 24.000 |
| 2 | 4.000 | 8.000 | 9.000 |
| 3 | 2.400 | 3.000 | No Upper Limit |
| a. | What is the optimal solution including the optimal value of the objective function? |
| b. | Suppose the profit on x1 is increased to $7. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $7? |
| c. | If the unit profit on x2 was $10 instead of $4, would the optimal solution change? |
| d. | If simultaneously the profit on x1 was raised to $5.5 and the profit on x2 was reduced to $3, would the current solution still remain optimal? |
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