Mathematics Part-II

Book: Mathematics Part-II

Chapter: 12. Linear Programming

Subject: Maths - Class 12th

Q. No. 9 of Exercise 12.2

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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.

let the diet contain x units of food F1 and y units of food F2.

x and y ≥ 0

The tabular representation of the data is

The cost of food F1 is Rs 4/unit and the cost of food F2 is Rs 6 per unit.

The constraints here are

3x + 6y ≥ 80

4x+ 3y ≥ 100

x and y ≥ 0

Total cost of the diet Z= 4x+ 6y

The mathematical formulation of the given data is

maximize Z= 4x+ 6y

Subject to constraints

3x + 6y ≥ 80

4x+ 3y ≥ 100

x and y ≥ 0

the feasible region by the system of constraints is as follows:

It can be seen the feasible region is unbounded with

The corner points of the feasible region are A(

The values of Z at the corner points are

As the feasible region is unbounded therefore 104 may or may not be the minimum value of Z.

For this we will draw a graph of inequality 4x+ 6y < 104

It can be seen that the feasible region has no common points with 4x+ 6y < 104

the maximum cost of the mixture will be Rs104.


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