Linear Programming — Class 12 Maths Solution

.: The system of constraints is :

$x + 3y \ge 3$ …(1)

$x + y \ge 2$ …(2)

and $x,y \ge 0$ …(3)

Let ${l_1}:x + 3y = 3$

${l_2}:x + y = 2$

The shaded region in the adjoining figure is the feasible region determined by the system of constraints (1) to (3).

The feasible region is unbounded.

We use Comer Point Method to determine the minimum value of Z.

We have :
$Z = 3x + 5y$ ...(4)

The co-ordinates of A, E and D are (3, 0), $\left( {\cfrac{3}{2},\cfrac{1}{2}} \right)$

(on solving $x + 3y = 3$ and $x + y = 2$) and (0, 2) respectively.

We evaluate Z at each corner point.

Now, since the region is unbounded we need to check whether 7 is the minimum value or not.

To decide this, we graph the inequality $3x + 5y < 7.$

Now, in the graph we observe 7 does not have points in common with feasible region.

So, 7 is the minimum value of Z.

Hence, ${Z_{\min }} = 7$ at $\left( {\cfrac{3}{2},\cfrac{1}{2}} \right)$