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Question
Solve the following Linear Programming Problems graphically:
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solution
The system of constraints is:
x + 3y ≥ 3 ....(i)
x + y ≥ 2 ....(ii)
and x, y ≥ 0 ....(iii)
Let l1 : x + 3y = 3
l2 : x + y = 2
The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iii).
The feasible region is unbounded.
We use the corner point method to determine the minimum value of Z,
We have,
Z = 3x + 5y
The co-ordinated of A, E and D are (3, 0), `(3/2, 1/2).`
(on solving x + 3y = 3 and x + y = 2) and (0, 2) respectively.
We evaluate Z at each corner point
| Corner point | Corresponding value of Z |
| (3, 0) | 9 |
| `(3/2, 1/2)` | 7 (Minimum) |
| (0, 2) | 10 |
Now, Since the region is unbounded we need to check whether 7 is the minimum value or not. To decide this, we graph the inequality 3x + 5y < 7.
Now, in the graph, we observe that 7 does not have points in common with a feasible region.
So, 7 is the minimum value at Z.
Hence, Zmin = 7 at `(3/2, 1/2)`
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