Let the diet contain x units of food and y units of food F2​. Therefore, x≥0 and y≥0The given information can be complied in a table as follows.The cost of food F1​ is Rs 4 per unit and of Food F2​ is Rs 6 per unit. Therefore, the constraints are3x+6y≥80 4x+3y≥100 Total cost of the diet, Z=4x+6y The mathematical formulation of the given problem isMinimise Z = 4x + 6 y .....(1)subject to the constraints,3x+6y≥80 .....(2)4x+3y≥100 .....(3)x,y≥0 ......(4)The feasible region determined by the constraints is as follows.It can be seen that the feasible region is unbounded.A(38​,0),B(2,21​), and C(0,211​)The corner points areA(380​,0),B(24,34​) and C(0,3100​)The values of Z at these corner points are as follows.As the feasible region is unbounded, therefore, 104 may or may not be the minimum, value of Z.