A firm manufactures two products A and B. Each product is processed on two machines M1 and M2. Product A requires 4 minutes of processing time on M1 and 8 min. on M2 ; product B requires 4 minutes on M1 and 4 min. on M2. The machine M1 is available for not more than 8 hrs 20 min. while machine M2 is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively.
Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.
Solution
Let x products of type A and y products of type B are manufactured.
Number of products cannot be negative.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as f
| Product |
\[M_1\]
|
\[M_2\]
|
| A(x) | 4 | 8 |
| B(y) | 4 | 4 |
| Availability | 500 | 600 |
Therefore, the constraints areollows:
\[4x + 4y \leq 500\]
\[8x + 4y \leq 600\]
The products A and B are sold at a profit of Rs 3 and Rs 4 respectively. Therefore, Profit gained from x products of type A and y products of type B is Rs 3x and Rs 4y respectively.
Total profit = Z = 3x + 4y which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[3x + 4y\]
subject to
\[4x + 4y \leq 500\]
\[8x + 4y \leq 600\]
First we will convert inequations into equations as follows:
4x + 4y = 500, 8x + 4y = 600, x = 0 and y = 0
Region represented by 4x + 4y ≤ 500:
The line 4x + 4y = 500 meets the coordinate axes at A1(125, 0) and B1(0, 125) respectively. By joining these points we obtain the line
4x + 4y = 500. Clearly (0,0) satisfies the 5x + 20y = 400 . So, the region which contains the origin represents the solution set of the inequation 5x + 20y ≤ 400.
Region represented by 8x + 4y ≤ 600:
The line 8x + 4y = 600 meets the coordinate axes at C1(75, 0) and D1(0, 150) respectively. By joining these points we obtain the line
8x + 4y = 600. Clearly (0,0) satisfies the inequation 8x + 4y ≤ 600. So,the region which contains the origin represents the solution set of the inequation
8x + 4y ≤ 600.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 4x + 4y ≤ 500, 8x + 4y ≤ 600, x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0), B1(0, 125), E1(25, 100) and C1(75, 0).
The values of Z at these corner points are as follows
| Corner point | Z= 3x + 4y |
| O | 0 |
| B1 | 500 |
| E1 | 475 |
| C1 | 225 |
The maximum value of Z is 500 which is attained at B1(0, 125).
Thus, the maximum profit is Rs 500 obtained when no units of product A and 125 units of product B were manufactured.
