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Questions
Solve the following Linear Programming Problems graphically:
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Consider the following Linear Programming Problem:
Minimise Z = x + 2y
Subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Show graphically that the minimum of Z occurs at more than two points.
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Solution 1
The system of constraints is:
2x + y ≥ 3 ....(i)
x + 2y ≥ 6 ....(ii)
and x ≥ 0, y ≥ 0 ...(iii)
Let l1 : 2x + y = 3
l2 : x + 2y = 6
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 is unbounded.
The coordinates of B and C are (0, 3) and (6, 0), respectively.
Applying the Corner Point Method, we have
| Corner point | Corresponding values of Z |
| (6, 0) | 6 |
| (0, 3) | 6 |
Since the region is unbounded, we need to check whether 6 is the minimum value or not. To decide this we graph the inequality x + 2y < 6.
Now, in the graph we observe 6 does not have points in common with the feasible region. So, 6 is the minimum value.
Hence Zmin = 6 at all points on the line segment joining the points (6, 0) and (0, 3).
Solution 2
The feasible region determined by the constraints 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0 is as shown.
The corner points of the unbounded feasible region are A(6, 0) and B(0, 3).
The values of Z at these corner points are as follows:
| Corner point | Value of the objective function Z = x + 2y |
| A(6, 0) | 6 |
| B(0, 3) | 6 |
We observe the region x + 2y < 6 has no points in common with the unbounded feasible region. Hence the minimum value of z = 6.
It can be seen that the value of Z at points A and B is the same. If we take any other point on the line x + 2y = 6, such as (2, 2) on line x + 2y = 6, then Z = 6.
Thus, the minimum value of Z occurs for more than 2 points and is equal to 6.
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