Linear Programming — Class 12 Maths Solution

.: The system of constraints is :

$x \ge 3$ …(1)

$x + y \ge 5$ ...(2)

$x + 2y \ge 6$ ...(3)

and $y \ge 0$ ...(4)

Let ${l_1}:x = 3;{l_2}:x + y = 5;{l_3}:x + 2y = 6;{l_4}:y = 0$

The shaded region in the figure is the feasible region determined by (1) to (4).

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

Applying Comer Point Method, we have

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.

$\therefore$ ${Z_{\max }} \ne 1$
Hence, Z has no maximum value.