Q. 35.0( 1 Vote )

To maintain one’s

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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