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 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?
Answer:
Let x and y are the number of bags of brand P and number of bags of brand Q respectively.
Now given problem can be converted into the mathematics form as
Minimize Z = 250x + 200y
subject to
3x + 3y/2 >= 18
=> 6x + 3y >=36
=> 2x + y >= 12...........1 (for element A)
5x/2 + 45y/4 >= 45
=> 10x + 45y >= 180
=> 2x + 9y >= 36.........2 (for element B)
2x + 3y >= 24 (for element C)
x >= 0, y >=0
Now the feasible region shows in the figure.
Point value of Z
(0, 12) 2400
(18, 0) 4500
(3,6) 1950
(9,2) 2650
Now value of Z is minimum at point (3,6).
So cost is minimum when 3 packets of brand P and 6 packets of brand Q are mixed.
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