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?
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 $z=250 x+200 y$. (1)
subject to the constraints,
$3 x+1.5 y \geq 18$ (2)
$2.5 x+11.25 y \geq 45$ (3)
$2 x+3 y \geq 24$ (4)
$x, y \geq 0$ (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 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, $250 x+200 y<1950$ or $5 x+4 y<39$, 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 $5 x+4 y<39$
Therefore, the minimum value of z is 1950 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.