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 nonnegativity restrictions. The shaded region in fig. represents the feasible region of the given LPP. The coordinates of the corner points of the feasible region areA_{2} (30, 0), P (10, 40), B_{1} (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

A_{2} (30, 0)
P (10, 40)
B_{1} (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.