Page 33 - DCOM303_DMGT504_OPERATION_RESEARCH

Operations Research

Notes Put x = 500 in eq. (1),
2
x + 2(500) = 2,000
1
Therefore x = 2,000 – 1,000
1
Therefore x = 1,000
1
At ‘C’: x + 2x = 2,000 ………(1)
1 2
x = 600 ………(2)
2
Put x = 600 in eq. (1),
2
x + 2(600) = 2,000
1
x = 2,000 – 1,200
1
Therefore x = 800
1
Step 5: Substitute the co-ordinates of corner points into the objective function.
Maximise ‘Z’ = 3x + 5x
1 2
At ‘O’, Z = 0 + 0 = 0

At ‘A’, Z = 3 (1,500) + 5 (0) = 4,500
At ‘B’, Z = 3 (1,000) + 5 (500) = 5,500

At ‘C’, Z = 3 (800) + 5 (600) = 5,400
At ‘C’, Z = 3 (0) + 5 (600) = 3,000

Inference

A maximum profit of ` 5,500 can be earned by producing 1,000 dolls of basic version and 500
dolls of deluxe version.

Example:

Maximise
‘Z’ = 2x + 3x (Subject to constraints)
1 2
x + x  400
1 2
2x + x  600
1 2
x , x  0 (Non-negativity constraints)
1 2
Solution:
Step 1: Find the divisible points on inequalities

Equation x1 x2
x1 + x2 = 400 400 400
2x1 + x2 = 600 300 600

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