A manufacturer of Furniture makes two products : chairs and tables. processing of these products is done on two machines A and B. A chair requires 2 hrs on machine A and 6 hrs on machine B. A table requires 4 hrs on machine A and 2 hrs on machine B. There are 16 hrs of time per day available on machine A and 30 hrs on machine B. Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Find with the help of graph what should be the daily production of each of the two products so as to maximize his profit.
Solution
Let x chairs and y tables were produced.
Number of chairs and tables cannot be negative.
Therefore, \[x, y \geq 0\]
The given information can be tabulated as follows:
Time on machine A(hrs) | Time on machine B (hrs) | |
Chairs | 2 | 6 |
Tables | 4 | 2 |
Availability | 16 | 30 |
Therefore, the constraints are
\[2x + 4y \leq 16\]
\[6x + 2y \leq 30\]
Profit gained by the manufacturer from a chair and a table is Rs 3 and Rs 5 respectively. Therefore, profit gained from x chairs and y tables is Rs 3x and Rs 5y.
Total profit = Z = \[3x + 5y\] which is to be maximised
Thus, the mathematical formulation of the given linear programmimg problem is
Max Z = \[3x + 5y\]
subject to
\[2x + 4y \leq 16\]
\[6x + 2y \leq 30\]
First we will convert inequations into equations as follows:
2x + 4y = 16, 6x + 2y =30, x = 0 and y = 0
Region represented by 2x + 4y ≤ 16:
The line 2x + 4y = 16 meets the coordinate axes at A1(8, 0) and B1(0, 4) respectively. By joining these points we obtain the line 2x + 4y = 16. Clearly (0,0) satisfies the 2x + 4y = 16. So, the region which contains the origin represents the solution set of the inequation 2x + 4y ≤ 16.
Region represented by 6x + 2y ≤ 30:
The line 6x + 2y =30 meets the coordinate axes at C1(5, 0) and D1(0, 15) respectively. By joining these points we obtain the line 6x + 2y =30 . Clearly (0,0) satisfies the inequation 6x + 2y ≤ 30. So,the region which contains the origin represents the solution set of the inequation 6x + 2y ≤ 30.
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 2x + 4y ≤ 16, 6x + 2y ≤ 30, x ≥ 0, and y ≥ 0 are as follows.The corner points are O(0, 0), B1(0, 4), E1
The values of Z at these corner points are as follows
Corner point | Z= 3x + 5y |
O | 0 |
B1 | 20 |
E1 | 22.2 |
C1 | 15 |
The maximum value of Z is 22.2 which is attained at B1