5 of 6 (6 complete) Consider the following LP problem developed at Zafar Malik\'

Question

5 of 6 (6 complete) Consider the following LP problem developed at Zafar Malik's Carbondale, Iilinois, optical scanning firm: 110 Maximize Z 1X,1x2 Subject to: 212 100 (C1) x,?20 On the graph on right, constraints C, and C2 have been ploted. C1 1x1 + 2x2 S 100 (C2) Isoprofit Line a) Using the point drawing tool, plot all the corner points for the feasible area. The optimum solution is: X1 = ? (round your response to two decimal places). 20 2round your response to two decimal places) Optimal solution value2+?(round your response to two decinalplaces). b) If a technical breakthrough occurred that raised the profit per unit of x1 to $3, due to this change, the optimal solution C2 0 10 20 30 40 50 60 TO 80 90 100 110 X1 Click the graph, choose a tool in the palette and follow the instructions to create your graph.

Explanation / Answer

Obective Function ,

Maximise ( 1*X1 + 1*X2)

Constraint :

2X1+X2=<100 --- C1

X1+2X2=< 100 ---- C2

X1,X2>=0

let

2X1+X2 = 100 ----- eq 1

X1+2X2= 100 -----eq2

2*eq1 - eq2

3x1 = 100

X1 = 100/3

X2 = 100/3

so point of intersection of two constraint is

100/3 & 100/3

Corner point under feasible area is (X1, X2)

(0,50), (50,0) & (100/3,100/3)

At (0,50)

Value of objective function = 50

At (50,0)

Value of objective function = 50

at (100/3,100/3)

Value of objective function =100/3+100/3 = 200/3 = 66.67

As we can see maximum value is at (100/3,100/3)

this is optimat solution

X1 = 100/3 = 33.33

X2 = 100/3 = 33.33

And Optimal Solution value = 1*100/3+1*100/3 = 200/3 = 66.67

2.

objective function = max(3*X1+ X2) -------- 1

Corner point under feasible area is (X1, X2)

(0,50), (50,0) & (100/3,100/3)

At (0,50)

New Value of objective function = 50

At (50,0)

New Value of objective function = 150

at (100/3,100/3)

New Value of objective function =100+100/3 = 400/3 = 133.33

As we can see maximum value is at (50,0)

so Optimal solution is (50,0)

X1 = 50

X2 = 0

Optimal solution value = 150

c) New objective function = 1.20*X1 + 1*X2

Corner point under feasible area is (X1, X2)

(0,50), (50,0) & (100/3,100/3)

At (0,50)

Z= 50

At (50,0)

Z = 60

At (100/3,100/3)

z = 120/3+100/3= 220/3 = 73.33

so maximum is at (100/3, 100/3)

X1 =100/3

X2= 100/3

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