MB0048 : A furniture manufacturer makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs 2 and Rs 10, respectively. What should be the daily production of each of the two products?
Posted October 20, 2011
on:MB0048 : A furniture manufacturer makes two products: chairs and tables. Processing of these products is done on two machines A and B. A chair requires 2 hours on machine A and 6 hours on machine B. A table requires 5 hours on machine A and no time on machine B. There are 16 hours per day available on machine A and 30 hours on machine B. Profit gained by the manufacturer from a chair and a table is Rs 2 and Rs 10, respectively. What should be the daily production of each of the two products?
Solution
Let x_{1 }= no of hours on machine A & x_{2 }=no of hours on machine B
Then the problem can be formulated as a P model as follows:-
Objective function, Maximise Z = 2x_{1} + 10x_{2}
Constraint equations: –
2x_{1} + 6x_{2} < 16
5x_{1 }< 30_{ }
x_{1}, x_{2} > 0
Step I
Treating all the constraints as equality, the first constraint is
2x_{1} + 6x_{2} < 16
Put x1 = 0 ⇒ = x_{2} = 2.7 ∴ The point is (0, 2.7)
Put x_{2} = 0 ⇒ = x_{1} = 8 ∴ The point is (8, 0) 2.7
Step II
Treating all the constraints as equality, the first constraint is
5x_{1 }< 30_{ }0_{ }6A_{ }8c
_{X1} = 6 ∴ The point is (6, 0)
Step III
The intersection of the above graphic denotes (A, B, C) the feasible region for the given problem.
Step IV
At corner points (O, A, B, C), find the profit value from the objective function. That point which maximize the profit is the optimal point.
Corner points |
W-Ordinates |
Objective Function Z = 2x_{1} + 10x_{2} |
Value |
A | (6, 0) | Z = 2(6) +0 | 12 |
B | (6, 2/3) | Z = 2(6) + 10 (2/3) | 19 |
C | (8, 0) | Z = 2(8) + 0 | 16 |
Result
The optimal solution is:
No of hours on machine A = 6
No of hours on machine B =2/3
Total profit, = 19 which is the maximum.
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