Let the mixture contain x kg. of food P and y kg. of food Q.
Clearly x ≥ 0, y ≥ 0.
Let Z be the total cost.
Table
Food |
Quantity (in units) |
Content of Vitamin A (in units) |
Content of Vitamin B (in units) |
Cost (in Rs.) |
P |
x |
3x |
5x |
60x |
Q |
y |
4y |
2y |
80y |
Total |
3x + 4y |
5x + 2y |
60x + 80y |
Mathematical formulation of the given problem is as follows:
Minimise Z = 60 x + 80 y
subject to the constraints
3x + 4y ≥ 8
5x + 2y ≥ 11
x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 3x + 4y = 8
For x = 0, 4y = 8 or y = 2
For y = 0, 3x = 8 or
and OY in L(0, 2)
Again we draw the graph of 5x + 2y = 11
For x = 0, 2y = 11 or
For y = 0, 5x = 11 or
and OY in
Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region and it is unbounded.
The corner points are
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph 60x + 80 y < 160 i.e. 3x + 4y < 8 ...(1)
Since (1) has no common point with feasible region.
minimum cost = 160 at
i.e. at points lying on segment joining
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F2 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.