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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.

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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.

Sum
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Solution

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 Zmax = 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.

∴ Zmax ≠ 1.

Hence, Z has no maximum value.

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Chapter 12: Linear Programming - Exercise 12.1 [Page 514]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 12 Linear Programming
Exercise 12.1 | Q 9 | Page 514

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