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Question
Solve the following linear programming problem graphically.
Maximize Z = 3x1 + 5x2 subject to the constraints: x1 + x2 ≤ 6, x1 ≤ 4; x2 ≤ 5, and x1, x2 ≥ 0.
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Solution
Since the decision variables, x1 and x2 are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
x1 + x2 = 6
| x1 | 0 | 6 |
| x2 | 6 | 0 |
x1 = 4 is a line parallel to x2-axis at a distance of 4 units.
x2 = 5 is a line parallel to x1-axis at a distance of 5 units.
The feasible region is OABCD and its co-ordinates are O(0, 0) A(4, 0) D(5, 0) and B is the point of intersection of the lines x1 + x2 = 6 and x1 = 4
Also C is the point of intersection of the lines x1 + x2 = 6 and x2 = 5
Verification of B:
x1 + x2 = 6 and x1 = 4
4 + x2 = 6
x2 = 2
∴ B is (4, 2)
Verification of C:
x1 + x2 = 6 and x2 = 5
x1 + 5 = 6
x1 = 1
∴ C is (1, 5)
| Corner points | Z = 3x1 + 5x2 |
| O(0, 0) | 0 |
| A(4, 0) | 12 |
| B(4, 2) | 22 |
| C(1, 5) | 26 |
| D(5, 0) | 15 |
Maximum of Z occurs at C(1, 5)
∴ The solution is x1 = 1, x2 = 5 and Zmax = 26.
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