Let the mixture contain x kg of food P and y kg of food Q. Therefore, x≥0 and y≥0The given information can be compiled in a table as follows.The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are3x+4y≥8 5x+2y≥11 Total cost, Z, of purchasing food is, Z=60x+80y The mathematical formulation of the given problem isMinimise Z=60x+80y…(1) subject to the constraints,3x+4y≥8… (2)5x+2y≥11… (3)x,y≥0…(4) The feasible region determined by the system of constraints is as follows.It can be seen that the feasible region is unbounded.The corner points of the feasible region are A(38​,0),B(2,21​), and C(0,211​).The values of Z at these corner points are as follows.As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.For this, we graph the inequality, 60 x+80 yIt can be seen that the feasible region has no common point with 3 x + 4y Therefore, the minimum cost of the mixture will be Rs 160 at the line segment joining the points (38​,0) and (2,21​).