A farmer mixes two brands P and Q of cattle feed. Brand P costing Rs.250 per bag contains 3 units of nutritional element A,2.5 units of element B and 2 units of element C. Brand Q costing Rs.200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A,B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag ? What is the minimum cost of the mixture per bag ?

.: Let 'x' bags of brand P and ‘y’ bags of brand Q of cattle feed be mixed.

We have,

The mathematical formulation of the problem is Minimize : $Z = 250x + 200y$ ...(1)

Subject to constraints
$3x + 1.5y \ge 18$ i.e., $2x + y \ge 12$ ...(2)

$2.5x + 11.25y \ge 45$ i.e. $2x + 9y \ge 36$ ...(3)

$2x + 3y \ge 24$ ….(4)

$x,y \ge 0$ …(5)

${l_1}:2x + y = 12$

${l_2}:2x + 9y = 36$

${l_3}:2x + 3y = 24$

Let us graph the inequalities (2) to (5).

The shaded portion represents the region which is unbounded.

Let us evaluate Z at the comer points C(18, 0), P(9, 2), Q(3, 6), B(0, 12).

In the table, we find that the smallest value of Z is 1950 at Q(3, 6).

Since the feasible region is unbounded, so we have to graph the inequality

$250x + 200y < 1950.$ i.e.

$5x + 4y < 39$
to check whether the resulting open half plane has point common with the feasible region.

From the graph, we see that it has no points in common.

Thus, the minimum value of Z is 1950 attained at point (3, 6).

Hence, 3 bags of brand P and 6 bags of brand Q of cattle feed are mixed for a minimum cost of Rs. 1950.