Determine graphically the minimum value of the objective function Z = – 50x + 20y ... (1)

subject to the constraints:

2x – y ≥ –5....(2)

3x + y ≥ 3…..(3)

2x – 3y ≤ 12...(4)

x ≥ 0, y ≥ 0.....(5) 

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded) is shown in the Figure. Observe that the feasible region is unbounded.

We now evaluate Z at the corner points. From this table, we find that –300 is the smallest value of Z at the corner point

(6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2).