Page 45 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes Therefore x = 22
2
Put x = 22 in eq. (1),
2
x + 3(22) = 90
1
x = 90 – 66
1
Therefore x = 24
1
At ‘C’: x + x = 40 …..(1)
1 2
x + 3x = 90 …..(2)
1 2
-2x = -50
2
Therefore x = 25
2
Put x = 25 in eq. (1),
2
x + 25 = 40
1
x = 15
1
Step 5: Substitute the co-ordinates of the corner points to the objective function
Maximise ‘Z’ = 40x + 60x
1 2
At ‘A’, Z = 40(30) + 60(10) = 1,800
At ‘B’, Z = 40(24) + 60(22) = 2,280

At ‘C’, Z = 40(15) + 60(25) = 2,100
Inference

Maximum profit can be obtained by producing 24 units of product A and 22 units of product B.

Example:
Maximise ‘Z’ = 7x + 3x (Subject to constraints)
1 2
x + 3x  3
1 2
x + x  4
1 2
x  5/2 or 2.5
1
x  3/2 or 1.5
2
x , x  0 (Non-negativing constraints)
1 2
Solution:
Step 1: Find the divisibles of the equalities

Equation x1 x2
x 1 + 3x 2 = 3 3 1
x1 + x2 = 4 4 4
x 1 = 2.5 2.5 0
x2 = 1.5 0 1.5

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