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प्रश्न
Solve the following Linear Programming Problems graphically:
Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
उत्तर
The system of constraints is
x + 2y ≤ 8 ....(i)
3x + 2y ≤ 12 ....(ii)
and x ≥ 0, y ≥ 0 ...(iii)
Let `l_1: x + 2y = 8; l_2: 3x + 2y = 12`
The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iii).
It is observed that the feasible region OCEB is bounded.
Thus, we use the Corner Point Method to determine the minimum value of Z.
We have, Z = - 3x + 4y
The co-ordinates of O, C, E and B are (0, 0), (4, 0), (2, 3)
(on solving x + 2y = 8 and 3x + 2y = 12) and (0, 4) respectively.
| Corner Point | Corresponding values of Z |
| (0, 0) | 0 |
| (4, 0) | -12 (Minimum) |
| (2, 3) | 6 |
| (0, 4) | 16 |
Hence, Zmin = -12 at the point ( 4, 0)
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