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Question
Solve the following linear programming problem graphically.
Minimize Z = 200x1 + 500x2 subject to the constraints: x1 + 2x2 ≥ 10; 3x1 + 4x2 ≤ 24 and x1 ≥ 0, 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 + 2x2 = 10
| x1 | 0 | 10 |
| x2 | 5 | 0 |
3x1 + 4x2 = 24
| x1 | 0 | 8 |
| x2 | 6 | 0 |
The feasible region is ABC and its co-ordinates are A(0, 5) C(0, 6) and B is the point of intersection of the lines
x1 + 2x2 = 10 ..........(1)
3x1 + 4x2 = 24 .........(2)
Verification of B:
3x1 + 6x2 = 30 ..........[(1) × 3]
3x1 + 4x2 = 24 .........(2)
− − −
2x2 = 6
x2 = 3
From (1), x1 + 6 = 10
x1 = 4
∴ B is (4, 3)
| Corner points | Z = 200x1 + 500x2 |
| A(0, 5) | 2500 |
| B(4, 3) | 2300 |
| C(0, 6) | 3000 |
Minimum value occurs at B(4, 3)
∴ The solution is x1 = 4, x2 = 3 and Zmin = 2300.
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