Answer :

let the diet contain x units of food F_{1} and y units of food F_{2}.

∴ x and y ≥ 0

The tabular representation of the data is

The cost of food F_{1} is Rs 4/unit and the cost of food F_{2} 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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