Letxbags of brand P andybags of brand Q should be mixed to produce the mixture.
Each bag of brand P costs250 and each bag of brand Q costs200. Therefore,xbags of brand P andybags of brand Q costs(250x+ 200y).
Since each bag of brand P contains 3 units of nutritional element A and each bag of brand Q contains 1.5 units of nutritional element A, therefore,xbag of brand P andybag of brand Q will contain (3x+ 1.5y) units of nutritional element A. But, the minimum requirement of nutrients A is 18 units.
3x+ 1.5y 18
2x+y 12
Similarly,xbag of brand P andybag of brand Q will contain (2.5x+ 11.25y) units of nutritional element B. But, the minimum requirement of nutrients B is 45 units.
2.5x+ 11.25y 45
2x+ 9y 36
Also,xbag of brand P andybag of brand Q will contain (2x+ 3y) units of nutritional element B. But, the minimum requirement of nutrients C is 24 units.
2x+ 3y 24
Thus, the given linear programming problem is
MinimiseZ= 250x+ 200y
subject to the constraints
2x+y 12
2x+ 9y 36
2x+ 3y 24
x,y 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6) and D(0, 12).
The value of the objective function at these points are given in the following table.
| Corner Point | Z= 250x+ 200y |
| (18, 0) | 250 18 + 200 0 = 4500 |
| (9, 2) | 250 9 + 200 2 = 2650 |
| (3, 6) | 250 3 + 200 6 = 1950 Minimum |
| (0, 12) | 250 0 + 200 12 = 2400 |
The smallest value ofZis 1950 which is obtained at(3, 6).
It can be seen that the open half-plane represented by250x+ 200y< 1950 or5x+ 4y< 39has no common points with the feasible region.
So, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost.
Hence, the minimum cost of the mixture per bag is 1950.
