Problems based on LPP when feasible region is unbounded.
1. Minimize z = 3x + 5y subject to constraints
X + y ≥ 2
X + 3y ≥ 3
X, y ≥ 0
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solution
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1. Let z = 3x + 5y
2. Let us Draw the line x + y = 6 and x + 3y = 3 using suitable points on the graph.
3.These lines intersect at
4. Now shade the region of intersection of these lines
5.Vertices of shaded feasible region are
Now,
At C(3,0)
At P(3\2,1\2)
At B(0,2)
Therefore at P(3\2,1\2), z=3x+5y is minimum
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P(3\2,1\2)
(3,0) P(3\2,1\2) and B(0,2)
Z= 3x + 5y
Z= 9
Z= 7
Z= 10
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