Determine graphically the minimum value of the objective function
Z = 50x + 20y
subject to the constraints:
2x – y ≥ – 3
3x + y ≥ 3
2x – 3y ≤ 12
x ≥ 0, y ≥ 0
A
50
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B
60
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C
100
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D
300
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Solution
The correct option is A 50 First of all, let us graph the feasible region of the system of inequalities.
Now evaluate Z at the corner points.
Corner Point
Z = 50x + 20y
(0, 3)
60
(1, 0)
50
(6, 0)
300
From this table, we find that the minimum value of Z is 50 which occure at the corner point (1,0) under the given constraints.