Let the factory produce x units of chemical A and y units of chemical B. Then the total cost is z = Rs (4x + 6y). This is the objective function which is to be minimized.
From the given table, the constraints are x + 2y ≥ 80, 3x + y ≥ 75.
Also, the number of units x and y of chemicals A and B cannot be negative.
∴ x ≥ 0, y ≥ 0.
∴ the mathematical formulation of given LPP is
Minimize z = 4x + 6y, subject to x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0.
First we draw the lines AB and CD whose equations are x + 2y = 80 and 3x + y = 75 respectively.
The feasible region is shaded in the graph. The vertices of the feasible region are A (80, 0), P and D (0, 75).
P is the point of intersection of the lines
x + 2y = 80 … (1)
and 3x + y = 75 … (2)
Multiplying equation (2) by 2, we get
6x + 2 y = 150
Subtracting equation (1) from this equation, we get
5x = 70
∴ x = 14
∴ from (2), 3(14) + y = 75
∴ 42 + y = 75
∴ y = 33
∴ P = (14, 33)
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(80)+ 6(0) =320 + 0
= 320 z(P) = 4(14)+ 6(33)
= 56+ 198 = 254 z(D)
= 4(0) + 6(75) = 0 + 450
= 450
∴ the minimum value of z is 254 at the point (14, 33).
Hence, 14 units of chemical A and 33 units of chemical B are to be produced in order to have the j minimum cost of Rs 254.
