Question

Solve The LP problem using the graphic method

Z Max=6X_{1}+5X_{2}

Constaint function:

X_{1} + 2X_{2} ≤ 240

3X_{1} + 2X_{2} ≤ 300

X_{1}≥ 0 , X_{2}≥0

Answer #1

Given data:

Objective function:

Max, Z=6X1+5X2

Constaint

X1 + 2X2 ≤ 240 ----------(1st constraint)

3X1 + 2X2 ≤ 300 ----------(2nd constraint)

X1≥ 0 , X2≥0

X1 intercepts in ghraph.

X1 = 240 in first constraint

X1 = 100 in second constraint

X2 intercepts in graphs.

X2 = 120 in 1st constraint

X2 =150 in second constraints

then the graph will be

the shaded area is the constraint satisfied area.

point A = (0,120)

point B =(30,105)

point C = (100,0)

These are the boundary points one of this point will give the maximum value.

Z= 6X1+5X2

Z(A) =(6*0)+5*120 = 600

Z(B) = 6*30 +5*105 =705

Z(C) = 6*100+ 5*0 = 600

So we are getting maximum value 705. at B point .

so the answer is

X1 =30

X2 = 105,

Max Z = 705.

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Constaint function:
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X1≥ 0 , X2≥0

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Constaint function:
2X1 + 4X2 ≤ 80
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X1≥ 0 , X2≥0

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X1, X2, X3 ≥ 0

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s.t. X1+2X2+X3=3
2X1-X2 =4 X1,X2,X3≥0
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Z
= 6X1 - 3X2
Subject to
2X1
+ 5X2 ≥ 10
3X1 + 2X2 ≤ 40
X1, X2 ≤ 15

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the problem is two dimensional, graph the feasible region, and
outline the progress of the algorithm.
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5X2
Subject to: X1 + 2X2 > (or equal to)
80
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X1, X2 > 0

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max 2x1+2x2+4x3
x1−2x2+2x3≤−1
3x1−2x2+4x3≤−3
x1,x2,x3≤0
Formulate a dual of this linear program. Select all the correct
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2. min −y1−3y2
3. y1+3y2≤2
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6. y1,y2≤0

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X1+2X2+ X3≥2
X1–X3≥1
X2+X3= 1
2X1+ X2≤3
X2, X3 ≥0, X1 urs
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