A firm is engaged in producing two products A and B. Each unit of product A requires two kg of raw material and four labour hours for processing whereas each unit of product B requires three kg of raw material and three hours of labour, of the same type. Every week, the firm has an availability of 60 kg of raw material and 96 labour hours. One unit of product A sold yields Rs. 40 and one unit of product B sold gives Rs. 35 as profit.
Formulate this problem as linear programming problem to determine as to how many units of each of the products should be produced per week so that the firm can earn the maximum profit. Assume that there is no marketing constraints so that all that is produced can be sold.

Let x and y represent the number of units produced per week of the products A and B respectively. Let Z be the profit corresponding to this rate of production. Then
Z = 40 x + 35 y ...(1)
In order to produce these number of units,
total consumption of raw material = 2x + 3y
and total labour hours needed = 4x + 3y
But total raw material available = 60 kg
and total labour hours available = 96
∴  we have,
                                             ...(2)
∵  it is not possible to produce negative number of units
∴ x ≥ 0, y ≥ 0    ...(3)
∴ firm’s allocation problem can be put in the following mathematical form:
Find two real numbers x and y such that
2x + 3y ≤ 60
4x + 3y ≤ 96
x ≥ 0, y ≥ 0
and for which the objective function
Z = 40x + 35y
is maximum
Consider a set of rectangular axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of
2x + 3y = 60

For x = 0, 3 y = 60 or y = 20
For y = 0, 2 x = 60 or x = 30
∴ line 2x + 3y = 60 meets OX in A (30, 0) and OY in B (0, 20).
Now we draw the graph of 4x + 3y = 96
For x = 0, 3y = 96 or y = 32
For y = 0, 4x = 96 or x = 24
∴ line 4x + 3y = 96 meets OX in C (24, 0) and OY in D (0, 32).
Since feasible region is the region which satisfies all the constraints.
∴  OCEB is the feasible region.
The corner points are O (0, 0), C (24, 0), E (18, 8), B (0, 20).
At O (0, 0), Z = 40 (0) 35 (0) = 0 + 0 = 0
At C (24, 0), Z = 40 (24) + 35 (0) = 960 + 0 = 960
At E (18, 8), Z = 40 (18)+ 35 (8) = 720 + 280 = 1000
At B(0, 20), Z = 40(0) + 35 (20) = 0 + 700 = 700
Here Rs. 1000 is the maximum values of Z and occurs at E (18, 8)
∴  optimal solution is x = 18, y = 8
i.e., 18 units of A and 8 units of B.