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.

Answer 1

The system of constraints is:

x ≥ 3                    ...(i)

x + y ≥ 5            ...(ii)

x + 2y ≥ 6        ...(iii)

and y ≥ 0          ...(iv)

Let l₁ : x = 3

l₂ : x + y = 5

l₃ : x + 2y = 6

l₄ : y = 0

The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iv).

The corner points are C(6, 0), E(4, 1) and F(3, 2).

Applying the Corner Point Method, we have

Corner Point	Corresponding values of Z
(6, 0)	-6
(4, 1)	-2
(3, 2)	1

It appears that Z_max = 1 at (3, 2).

But the feasible region is unbounded, therefore, we draw the graph of the inequality -x + 2y > 1.

Since the half-plane represented by - x + 2y > 1 has points common with the feasible region.

∴ Z_max ≠ 1.

Hence, Z has no maximum value.