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 F2 costs ₹ 4 per unit F2 costs ₹ 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 and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.

Question

Philoid · Accepted Answer

The above information can be expressed in the following table:

Let the quantity of the foods F1 and F2 be ‘x’ and ‘y’ respectively.

Cost of food F1 = 4x

Cost of food F2 = 6y

Cost of Diet = 4x + 6y

Now,

⟹ 3x + 6y ≥ 80

i.e. the minimum requirement of Vitamins from the two foods is 80units, each of which contains 3units and 6units of Vitamins.

⟹ 4x + 3y ≥ 100

i.e. the minimum requirement of minerals firm the two foods is 100units, each of which contains 4unit and 3 units of vitamins.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and ‘y’ that:

Minimise Z = 4x + 6y

Subject to the following constraints:

(i) 3x + 6y ≥ 80

(ii) 4x + 3y ≥ 100

(iii) x,y ≥ 0 (∵ quantity cant be negative)

The feasible region is unbounded.

The corner points of the feasible region is as follows:

Z is smallest at

Let us consider 4x + 6y ≤ 104

As it has no intersection with the feasible region, the smallest value is the minimum value.

The minimum cost of diet is ₹104.