Page 38 - DCOM303_DMGT504_OPERATION_RESEARCH

Unit 2: Linear Programming Problems

Therefore 10.5 + x = 45 Notes
2
x = 34.5
2
At ‘B’: x = 20,
1
Therefore 3x + x = 66
1 2
Therefore 3(20) + x = 66
2
Therefore x = 66 – 60
2
Therefore x = 6
2
Step 5: Substituting the co-ordinates of corner points into objective function.
Maximise ‘Z’ = 8000x + 7000x
1 2
At ‘O’, Z = 8000(0) + 7000(0) = 0

At ‘A’, Z = 8000(20) + 7000(0) = 1,60,000
At ‘B’, Z = 8000(20) + 7000(6) = 2,02,000

At ‘C’, Z = 8000(10.5) + 7000(34.5) = 3,25,500
At ‘D’, Z = 8000(5) + 7000(40) = 3,20,000

At ‘E’, Z = 8000(0) + 7000(40) = 2,80,000

Inference

To maximize the profit, i.e., at ` 3,25,500 the company has to manufacture 10,500 bottles of type
A medicine and 34,500 bottles of type B medicine.

Task Give the graphical solution for the following LPP
Maximize ‘Z’ = 0.50x – 0.10x , (Subject to constraints)
2 1
2x + 5x  80
1 2
x + x  20
1 2
x , x  0
1 2

!
Caution It is very much essential to locate the solution point of the LPP with respect to the
objective function type (max or min). If the given problem is maximization, zmax then
locate the solution point at the far most point of the feasible zone from the origin and if
minimization, Zmin then locate the solution at the shortest point of the solution zone
from the origin.

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