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  • A diet is to contain at least 80 units of vitamin A and 100 units of minerals.Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

9.  A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

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Let  diet contain x unit of food F1  and y unit of foof F2  .Thus, x\geq 0,y\geq 0.

The given information can be represented in table as :

  Vitamin  minerals  cost per unit 
foof F1   3     4       4
food F2    6      3        6
      80     100  

 Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint  are  

                3x+4y\geq 4

               6x+3y\geq 6

                x\geq 0,y\geq 0

                 Z= 4x+6y

The feasible  region determined by constraints is as follows:

    

We can see feseable region is unbounded.

The corner points of feasible region are  A(\frac{80}{3},0),B(24,\frac{4}{3}),C(0,\frac{100}{3})

The value of Z at corner points is as shown :

Corner points

Z= 4x+6y

  
A(\frac{80}{3},0)             106.67  
B(24,\frac{4}{3}),               104 minimum
C(0,\frac{100}{3})                 200 maximum
                        

  Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw 4x+6y< 104   or   2x+3y< 52  and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with  2x+3y< 52.

Hence , Z has minimum value 104  .

Posted by

seema garhwal

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