Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0. Show that the minimum of Z occurs at more than two points.

It is given in the question that,

Z = x + 2y

We have to subject on the following equation:

(x, y) = (0, 3), (6, 0)

(x, y) = (0, 3), (1, 1)

As per the table minimum value of Z is 6 but we can see that the feasible region is unbounded. Thus, 6 may or may not be the minimum value of Z

∴ We will plot the graph of inequality,

Here we will see that there is no common point with the feasible region.

Hence, Z will be minimum on all points joining line (0, 3), (6, 0). Therefore, Z will be minimum on x + 2y = 6

6