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
⟹ 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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