The given data may be put in the following tabular form:
Supplier
Chemical
|
S
|
T
|
Minimum Requirement
|
A
B
|
4
2
|
1
1
|
80
60
|
Cost per unit
|
Rs 10
|
Rs 4
|
|
Suppose x units of mix are purchased from supplier S and y units per purchased from supplier T.
Total cost Z = 10x + 4y.
Units of chemical A per bottle = 4x + y.
But the minimum requirement of chemical A per bottle = 80
∴ 4x + y ≥ 80
Similarly, 2x + y ≥ 60
Clearly, x ≥ 0, y ≥ 0
Thus, the mathematical formulation of the given LPP is
Minimize Z = 10x + 4y
Subject to
4x + y ≥ 80
2x + y ≥ 60
and, x ≥ 0, y ≥ 0
Now, we find the feasible region which is the set of all points whose coordinates simultaneously satisfy all constraints including non-negativity restrictions. The shaded region in fig. represents the feasible region of the given LPP. The coordinates of the corner points of the feasible region areA2 (30, 0), P (10, 40), B1 (0, 80).
These points are obtained by solving the equations of the corresponding intersecting lines, simultaneously.
The values of the objective function at these points are given in the following table:
Point (x, y)
|
Value of objective function
Z = 10x + 4y
|
A2 (30, 0)
P (10, 40)
B1 (0, 80)
|
Z = 10 × 30 + 4 × 0 = 300
Z = 10 × 10 + 40 × 4 = 260
Z = 10 × 0 + 4 × 80 = 320.
|
Clearly, Z is minimum at (10, 40). The feasible region is unbounded and the open half plane represented by 10x + 4y < 260 does not have points in common with the feasible region. So, Z is minimum at x = 10, y = 40. Hence, x = 10, y = 40 is the optimal solution of the given LPP.
Hence, the cost per bottle is minimum when the company purchases 10 mixes from supplier S and 40 mixes from supplier T.