Answer :

Let the quantity of foods chosen be ‘x’ and ‘y’

Cost of food X = 0.6x

Cost of food Y = y

Cost of diet = 0.6x + y

Now,

⟹ 10x + 4y ≥ 20

i.e. the minimum daily requirement of calcium in the diet is 20 units.

⟹ 5x + 6y ≥ 20

i.e. the minimum daily requirement of protein in the diet is 20 units.

⟹ 2x + 6y ≥ 12

i.e. the minimum daily requirement of calories in the diet is 12 units.

Hence, mathematical formulation of the LPP is as follows:

Find ‘x’ and y’ such that

Minimises Z = 0.6x + y

Subject to the following constraints:

(i) 10x + 4y ≥ 20

(ii) 5x + 6y ≥ 20

(iii) 2x + 6y ≥ 12

(iv) 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 0.6x + y ≤ 2.712.

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

The minimum value of Z is ₹2.712

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