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Sample Questions in Graphical Solution Procedure

Solved problems in Graphical Solution Procedure, sample assignments and homework

Questions: Minimize Z = 10x1 + 4x2

Subject to

3x1 + 2x2 ≥ 60

            7x1 + 2x2 ≥ 84

            3x1 +6x2 ≥ 72

x1 ≥ 0 , x2 ≥ 0

 

Answer

The first constraint 3x1 + 2x2 ≥ 60, can be written in form of equation

3x1 + 2x2 = 60

Place x1 =0, then x2 = 30

Place x2 =0, then x1 = 20

Then the coordinates are (0, 30) and (20, 0)

 

The second constraint 7x1 + 2x2 ≥ 84, can be written in form of equation

7x1 + 2x2 = 84

Place x1 =0, then x2 = 42

Place x2 =0, then x1 = 12

The coordinates then are (0, 42) and (12, 0)

 

The third constraint 3x1 +6x2 ≥ 72, can be written in form of equation

3x1 +6x2 = 72

Place x1 =0, then x2 = 12

Place x2 =0, then x1 = 24

Thus, coordinates are (0, 12) and (24, 0)

 

The graphical presentation is

 

 1485_Graphical Solution Procedure Sample Assignment.png 

 

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 7x1 + 2x2 = 84 and 3x1 + 2x2 = 60 to obtain the coordinates)

C (18, 3) Solve the two equations 3x1 +6x2 = 72 and 3x1 + 2x2 = 60 to obtain the coordinates)

D (24, 0)

 

We are given that Min Z = 10x1 + 4x2

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 x1 = 6, x2 = 21

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