Question

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

Minimise and Maximise Z = x + 2y 

subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

Answer 1

The system of constraints is:

x + 2y ≥ 100           ....(i)

2x - y ≤ 0                ....(ii)

2x + y ≤ 200           ....(iii)

and x, y ≥ 0           ....(iv)

Let l₁ : x + 2y = 100

l₂ : 2x - y = 0

l₃ : 2x + y = 200

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

It is observed that the feasible region ECDB is bounded.

Thus, we use the Corner Point Method to determine the maximum and minimum values of Z.

We have, Z = x + 2y

The co-ordinates of E, C, D and B are (20, 40) (on solving x + 2y = 100 and 2x - y = 0), 

(50, 100) (on solving 2x + y = 200 and 2x - y = 0), (0, 200) and (0, 50) respectively.

Corner Point	Corresponding values of Z
(20, 40)	100
(50, 100)	250
(0, 200)	400 (Maximum)
(0, 50)	100

Hence Z_max = 400 at (0, 200) and Zmin = 100 at all points on the line segment joining the points (0, 50) and (20, 40).