Question

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
• B. 60
• C. 100
• D. 300

Answer 1

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.