# L2 2 0x 1 5 x 2 40 20 x 1 60 x 2 200 4 x 1 x 2 10 0 0

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L2:20X1+ 5 X2=40-20X1- 60 X2= -200(-4)X1X210003.3X1X22008X1X240030
20X1+ 5 X2=40-55 X2= - 160X2= - 160/-55X2= 2.91X2en L15X1+ 15 X2= 505X1=50- 15 X2X1=10-3 X2X1=10-3(2.91)X1=1.27Punto C L1 y L3L1:5X1+ 15 X2= 50(-3)L3:15X1+ 2 X2= 60-15X1- 45 X2= -150(-4)15X1+ 2 X2=60-43 X2= - 90X2= - 90/-43X2= 2.09X2en L315X1+ 2 X2= 6015X1=60- 2 X2X1=3.726.- Punto ÓptimoA) Z (0,8) = 4(0)+2(8) = 16B) Z (1.27, 2.91)= 4(1.27)+2(2.91) = 10.9C) Z (3.72, 2.09)=4(3.72)+2(2.09) = 20.68D) Z (0,30) = 4(0)+2(30) = 607.- ComprobaciónPunto optimo
5 x1+15 x2505 (1.27) +15 (2.91)≥ 50505020 x1 + 5 x2 ≥ 4020 (1.27) + 5 (2.91) ≥ 4039.95 ≥ 4015 x1 + 2 x2 ≤ 6015 (1.27)+ 2 (2.91)≤ 6024.87 ≤ 60INTERPRETACION:Se deben consumir 1.27 porciones de res y 2.91 de papas a un costo mínimo de\$ 10.9
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