Question

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 elements 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?

Answer 1

Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows.

The given problem can be formulated as follows.
Minimize z = 250x + 200y … (1)
subject to the constraints,

The feasible region determined by the system of constraints is as follows.

The corner points of the feasible region are A (18, 0), B (9, 2), C (3, 6), and D (0, 12).
The values of z at these corner points are as follows.

As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of z.

For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 5x + 4y < 39
Therefore, the minimum value of z is 2000 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950.