Answer :

let the diet contain x units of food F1 and y units of food F2.

x and y ≥ 0


The tabular representation of the data is



The cost of food F1 is Rs 4/unit and the cost of food F2 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.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A manufacturer prMathematics - Board Papers

A retired person Mathematics - Board Papers

A manufacturing cMathematics - Board Papers