#### Question

Show that the minimum of Z occurs at more than two points

Maximise Z = – x + 2y, subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

#### Solution

The feasible region determined by the constraints, x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0. is as follows.

It can be seen that the feasible region is unbounded.

The values of Z at corner points A (6, 0), B (4, 1), and C (3, 2) are as follows

Corner point |
Z = −x + 2y |

A(6, 0) | Z = − 6 |

B(4, 1) | Z = − 2 |

C(3, 2) | Z = 1 |

As the feasible region is unbounded, therefore, Z = 1 may or may not be the maximum value.

For this, we graph the inequality, −*x* + 2*y* > 1, and check whether the resulting half plane has points in common with the feasible region or not.

The resulting feasible region has points in common with the feasible region.

Therefore, Z = 1 is not the maximum value. Z has no maximum value