Advertisements
Advertisements
प्रश्न
Solve the following linear programming problem graphically.
Maximise Z = 4x1 + x2 subject to the constraints x1 + x2 ≤ 50; 3x1 + x2 ≤ 90 and x1 ≥ 0, x2 ≥ 0.
Advertisements
उत्तर
Maximise Z = 4x1 + x2
Subject to the constraints
x1 + x2 ≤ 50
3x1 + x2 ≤ 90
x1, x2 ≥ 0
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 = 50
| x1 | 0 | 50 |
| x2 | 50 | 0 |
3x1 + x2 = 90
| x1 | 0 | 30 |
| x2 | 90 | 0 |
The feasible region is OABC and its co-ordinates are O(0, 0) A(30, 0) C(0, 50) and B is the point of intersection of the lines
x1 + x2 = 50 ..........(1)
and 3x1 + x2 = 90 .........(2)
Verification of B:
x1 + x2 = 50 ..........(1)
3x1 + x2 = 90 .........(2)
− − −
− 2x1 = − 40
x1 = 20
From (1), 20 + x2 = 50
x2 = 30
∴ B is (20, 30)
| Corner points | Z = 4x1 + x2 |
| O(0, 0) | 0 |
| A(30, 0) | 120 |
| B(20, 30) | 110 |
| C(0, 50) | 50 |
Maximum value occurs at A(30, 0)
Hence the solution is x1 = 30, x2 = 0 and Zmax = 120.
