A producer of furniture manufactures two products  tables and chairs. Processing of these products is done on two machines A and B. A chair needs 2 hours on machine A and 6 hours on machine B. A table needs 5 hours on machine A and no time on machine B. There are 16 hours of time per day accessible on machine A and 30 hours on machine B. Profit earned by the manufacturer from a chair and a table is Rs 2 and Rs 10 correspondingly. What must be the everyday production of each of two products?
Answer
Assume x_{1} indicates the number of chairs
Assume x_{2} indicates the number of tables

Chairs

Tables

Availability

Machine A
Machine B

2
6

5
0

16
30

Profit

Rs 2

Rs 10


LPP
Max Z = 2x_{1} + 10x_{2}
Subject to
2x_{1}+ 5x_{2 }≤ 16
6x_{1} + 0x_{2} ≤ 30
x_{1 }≥ 0 , x_{2 }≥ 0
Solve graphically
The first constraint 2x_{1}+ 5x_{2 }≤ 16, can be written in the form of equation
2x_{1}+ 5x_{2 }= 16
Place x_{1} = 0, then x_{2} = 16/5 = 3.2
Place x_{2} = 0, then x_{1} = 8
The coordinates are (0, 3.2) and (8, 0)
The second constraint 6x_{1} + 0x_{2} ≤ 30, can be written in the form of equation
6x_{1} = 30 → x_{1} =5
The corner positions of feasible region are A, B and C. So the coordinates for the corner positions are
A (0, 3.2)
B (5, 1.2) (Solve the two equations 2x_{1}+ 5x_{2 }= 16 and x_{1} =5 to obtain the coordinates)
C (5, 0)
We are given that Max Z = 2x_{1} + 10x_{2}
At A (0, 3.2)
Z = 2(0) + 10(3.2) = 32
At B (5, 1.2)
Z = 2(5) + 10(1.2) = 22
At C (5, 0)
Z = 2(5) + 10(0) = 10
Max Z = 32 and x_{1} = 0, x_{2} = 3.2
The manufacturer must manufacture about 3 tables and no chairs to obtain the max profit.