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Question
Solve the following Linear Programming Problems graphically:
Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.
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Solution
The system of constraints is
x + 2y ≤ 10 ....(i)
3x + y ≤ 15 ....(ii)
and x, y ≥ 0 ....(iii)
Let l1 : x + 2y = 10
l2 : 3x + y =15
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 maximum value of Z.
We have: Z = 3x + 2y ....(iv)
The coordinates of O, C, E and B are (0, 0) (5, 0), (4, 3) (solving x + 2y = 10, 3x + y = 15) and (0, 5) respectively.
| Corner point | Corresponding values of Z |
| (0, 0) | 0 |
| (5, 0) | 15 |
| (4, 3) | 18 (Maximum) |
| (0, 5) | 10 |
Hence Zmax = 18 at (4, 3).
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