Solved problems in Graphical Solution Procedure, sample assignments and homework
Questions: Minimize Z = 10x_{1} + 4x_{2}
Subject to
3x_{1} + 2x_{2} ≥ 60
7x_{1} + 2x_{2} ≥ 84
3x_{1} +6x_{2 }≥ 72
x_{1 }≥ 0 , x_{2 }≥ 0
Answer
The first constraint 3x_{1} + 2x_{2} ≥ 60, can be written in form of equation
3x_{1} + 2x_{2} = 60
Place x_{1} =0, then x_{2} = 30
Place x_{2} =0, then x_{1} = 20
Then the coordinates are (0, 30) and (20, 0)
The second constraint 7x_{1} + 2x_{2} ≥ 84, can be written in form of equation
7x_{1} + 2x_{2} = 84
Place x_{1} =0, then x_{2} = 42
Place x_{2} =0, then x_{1} = 12
The coordinates then are (0, 42) and (12, 0)
The third constraint 3x_{1} +6x_{2 }≥ 72, can be written in form of equation
3x_{1} +6x_{2 }= 72
Place x_{1} =0, then x_{2} = 12
Place x_{2} =0, then x_{1} = 24
Thus, coordinates are (0, 12) and (24, 0)
The graphical presentation is
The corner positions of feasible region are A, B, C and D. Thus the coordinates for the corner points are
A (0, 42)
B (6, 21) (Solve the two equations 7x_{1} + 2x_{2} = 84 and 3x_{1} + 2x_{2} = 60 to obtain the coordinates)
C (18, 3) Solve the two equations 3x_{1} +6x_{2 }= 72 and 3x_{1} + 2x_{2} = 60 to obtain the coordinates)
D (24, 0)
We are given that Min Z = 10x_{1} + 4x_{2}
At A (0, 42)
Z = 10(0) + 4(42) = 168
At B (6, 21)
Z = 10(6) + 4(21) = 144
At C (18, 3)
Z = 10(18) + 4(3) = 192
At D (24, 0)
Z = 10(24) + 4(0) = 240
The minimum value is calculated at the point B. Consequently Min Z = 144 and x_{1} = 6, x_{2} = 21