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  • 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 Determine the number of bags of each brand which should be mixed in order

2.  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?

Answers (1)

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Let farmer mix x  bags of brand  P and y bags of brand Q. Thus, x\geq 0,y\geq 0.

The given information can be represented in the table as :

  Vitamin A  Vitamin B  Cost 
Food P   3     5     60
Food Q    4      2      80
 requirement     8     11  

The given problem can be formulated as follows:

Therefore, we have 

 3x+1.5y\geq 18

 2.5x+11.25y\geq 45

 2x+3y\geq 24

 Z=250x+200y

Subject to constraint,

 3x+1.5y\geq 18

 2.5x+11.25y\geq 45

 2x+3y\geq 24

  x\geq 0,y\geq 0

The feasible  region determined by constraints is as follows:

    

The corner points of the feasible region are A(18,0),B(9,2),C(3,6),D(0,12)

The value of Z at corner points is as shown :

 corner points 

Z=250x+200y

  
   A(18,0)             4500  

B(9,2)

             2650  
C(3,6)               1950 minimum
 D(0,12)               2400  

Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw 250x+200y< 1950 and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with  250x+200y< 1950.

Hence, Z has a minimum value 1950  at point C(3,6).

Posted by

seema garhwal

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