Maximizar método simplex
El fabricante dispone para la confección de 750 m de tejido de algodón y 1000 m de tejido de poliéster. Cada pantalón precisa 1 m de algodón y 2 m de poliéster. Para cada chaqueta se necesitan 1.5 m de …ver más…
) + (200) <=1000
400+200 < = 1000
600 <= 1000
VERTICES FUNCION OBJETIVO Z= 50X + 40Y
(0, 0) 50(0) + 40(0)=0
(0, 500) 50(0) + 40(500) = 20,000
(500, 0) 50(500) + 40(0) = 25,000
(375, 250) 50(375) + 40(250) = 28,750
RESPUESTA:
La solución optima es fabricar 375 pantalones y 250 chaquetas para obtener un beneficio de 28,750€
| | METODO SIMPLEX TABULAR | | | | | | | | | | | | Max | | | | | | | Z = 50 X + 40Y | | | | | | | RESTRCIONES | | | | | | | 2X + 3Y<=1500 | | | | | | | 2X + Y <=1000 | | | | | | | Holguras | | | | | | | X3 | | | | | | | | X4 | | | | | | | | | Z | X | Y | X3 | X4 | LD | | 0 | 1 | -50 | -40 | 0 | 0 | 0 | | X3 | 0 | 2 | 3 | 1 | 0 | 1500 | | X4 | 0 | 2 | 1 | 0 | 1 | 1000 | | | | | | | | | | | Z | X | Y | X3 | X4 | LD | | 0 | 1 | 0 | -15 | 0 | 25 | 25,000 | | X3 | 0 | 0 | 2 | 1 | -1 | 750 | | X4 | 0 | 1 | 1/2 | 0 | 1/2 | 500 | | | | | | | | | | | Z | X | Y | X3 | X4 | LD | | 0 | 1 | 0 | -15 | 0 | 25 | 25,000 | | X3 | 0 | 0 | 2 | 1 | -1 | 750 | | X | 0 | 0 | 1 | 0 | 1/2 | 500 | | | | | | | | | | | Z | X | Y | X3 | X4 | LD | | 0 | 1 | 0 | 0 | 7 1/2 | 35/2 | 28,500 | | X3 | 0 | 0 | 1 | 1/2 | - 1/2 | 375 | | X | 0 | 0 | 0 | - 1/2 | 1 | 250 | | | | | | | | | | | Z | X | Y | X3 | X4 | LD | | 0 | 1 | 0 | 0 | 7 1/2 | 35/2 |