Question

A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs250 per bag contains 3 units of nutritional element A,2.5units of element B and 2units of element C.Brand Q costing Rs200 per bag contains 1.5units of element B and 3units 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 that 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,
3x+1.5y≥18.....(2)
2.5x+11.25y≥45....(3)
2x+3y≥24....(4)
x,y≥0.....(5)

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 value of Z at these corner points is 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 for 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.