The given information can be summarized in the following tabular form:
Machines
|
Time required to produce products
|
Max. machine hours available
|
Nut
|
Bolt
|
A
B
|
1
3
|
3
1
|
12
12
|
Profit (in Rs)
|
2.50
|
1.00
|
|
Let the manufacturer produce x packages of nuts and y packages of bolts each day. Since machine A takes one hour to produce one package of nuts and 3 hours to produce one package of bolts. Therefore, the total time required by machine A to produce x packages of nuts and y packages of bolts is (x + 3y) hours. But machine A operates for at most 12 hours.
∴ x + 3y ≤ 12
Similarly, the total time required by machine B to produce x packages of nuts and y packages of bolts is (3x + y) hours. But machine B operates for at most 12 hours.
∴ 3x + y ≤ 12
Since the profit on one package of nuts is Rs 2.50 and on one package of bolts the profit is Rs. 1. Therefore, profit on x packages of nuts and ypackages of bolts is of Rs (2.50x + y). Let Z denote the total profit. Then,Z = 2.50x + y.
Clearly, x ≥ 0 and y ≥ 0
Thus, the above LPP can be stated mathematically as follows:
Maximize Z = 2.50x + y
Subject to x + 3y ≤ 12
3x + y ≤ 12
and, x, y ≥ 0
To solve this LPP graphically, we first convert the inequations into equations to obtain the following equations.
x + 3y = 12, 3x + y = 12, x = 0, y = 0
The line x + 3y = 12 meets the coordinate axes at A1 (12, 0) and B1 (0, 4). Join these two points to obtain the line represented by x + 3y = 12. The region represented by the inequation x + 3y ≤ 12 is the region containing the origin as x = 0, y = 0 satisfies the inequation x + 3y ≤ 12.
The line 3x + y = 12 meets the coordinate axes at A2 (4, 0) and B2 (0, 12). Join these points to obtain the line represented by 3x + y = 12. Since x = 0,y = 0 satisfies the inequation 3x + y ≤ 12. So, the region containing the origin and below the line 3x + y = 12 represents the region represented by 3x + y ≤ 12.
Clearly, x ≥ 0 and y ≥ 0 represent all points in first quadrant.
Thus, the shaded region OA2 PB1 in fig. represents the feasible region of the given LPP.
The coordinates of the corner-points of the feasible region OA2PB1 are O(0, 0). A2(4, 0), P(3, 3) and B1 (0, 4). These points are obtained by solving the corresponding intersecting lines simultaneously.
The values of the objective function at the corner-points of the feasible region are given in the following table:
Point (x, y)
|
Value of the objective function
Z = 2.50x + y
|
O (0, 0)
A2 (4, 0)
P (3, 3)
B1 (0, 4)
|
Z = 2.50 × 0 + 1 × 0 = 0
Z = 2.50 × 4 + 1 × 0 = 10
Z = 2.50 × 3 +1 × 3 = 10.50
Z = 2.50 × 0 + 1 × 4 = 4
|
Clearly, Z is maximum at x = 3, y = 3 and the maximum value of Z is 10.50.
Hence, the optimal production strategy for the manufacturer will be to manufacture 3 packages each of nuts and bolts daily and in this case his maximum profit will be Rs 10.50