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20 06

The document presents a linear programming model to maximize the objective function Z= 50X1 + 56X2 subject to the constraints X1 + X2 <= 80, 3X1 + 2X2 <= 220, and 2X1 + 3X2 <= 210, with non-negativity conditions on X1 and X2. By solving the model using linear algebra techniques, it identifies five feasible solutions: (0,0), (0,70), (30,50), (60,20), and (73.33,0). It then evaluates the objective function Z at each feasible point, determining that the optimal solution is (30,50), with a maximum value of Z = 4300.

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SAUL CRUZ ECHEVERRIA
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32 views2 pages

20 06

The document presents a linear programming model to maximize the objective function Z= 50X1 + 56X2 subject to the constraints X1 + X2 <= 80, 3X1 + 2X2 <= 220, and 2X1 + 3X2 <= 210, with non-negativity conditions on X1 and X2. By solving the model using linear algebra techniques, it identifies five feasible solutions: (0,0), (0,70), (30,50), (60,20), and (73.33,0). It then evaluates the objective function Z at each feasible point, determining that the optimal solution is (30,50), with a maximum value of Z = 4300.

Uploaded by

SAUL CRUZ ECHEVERRIA
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Modelo de PL

Max Z= 50X1 + 56X2


s.a.
X1 + X2 <= 80
3X1 + 2X2 <= 220
2X1 + 3X2 <= 210
Toda X1 y X2 >= 0

X1 + x2 = 80 3X1 + 2x2 = 220 2X1 + 3x2 = 210

X1 = 0 X1 = 0 X1 = 0

0 + x2 = 80 3(0) + 2x2 = 220 2(0) + 3x2 = 210

X2 = 80 2X2 = 220 3X2 = 210

X2 = 0 X2 = 220/2 X2 = 210/3

X1 + 0 = 80 X2 = 110 X2 = 70

X1 = 80 X2 = 0 X2 = 0

(80,80) 3X1 + 2(0) = 2X1 + 3(0) = 210


220
2X1 = 210
3X1 = 220
X1 = 210/2
X1 = 220/3
X1 = 105
X1 = 73.33
(105,70)
(73.33,110)
A (0,0) X1 + X2 = 80 + 2X1 + 3X2 = 210 3X1 + 2X2 = 220 + X1 + X2 = 80
B (0,70) -3X1 -3X2 = -240 + 2X1 +3X2 = 210 3X1 + 2X2 = 220
C R1-R3 -1X1 + 0X2 =-30 -3X1 + -3X2 =-240
D R2-R1 -1X1 = -30 0X1 –X2 = -20
E (73.33,0) X1 = -30/-1 X2 = 20
X1 = 30
3X1 + 2(20) = 220
2(30) + 3X2 = 210 3X1 + 40 = 220
60 + 3X2 = 210 3X1 = 220 – 40
3X2 = 210 -30 3X1 = 180
3X2 =150 X1 = 180/3
X2 = 150/3 X1 = 60
X2 = 50
(60,20)
(30,50)

A (0,0) Modelo de PL 30 + 50 = 80
B (0,70) Max Z: 50 X1 + 56 X2 30(30) + 2(50) = 190
C R1-R3 s.a. 2(30) + 3(50) = 210
D R2-R1 X1 + X2 <= 80
E (73.33,0) 3X1 + 2X2 <= 80
2X1 + 3X2 <= 210
Toda X1 y X2 >= 0

Max Z= 50X1 + 56X2


A Z= 50(0) + 56(0) = 0
B Z = 50(0) + 56(70)= 3920
C Z = 50(30) + 56(50) = 4300
D Z = 50(60) + 56(20) = 4120
E Z = 50(73.33) + 56(0) = 3666.5

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