Linear programming problem (50 points) A manufacturer produces pens and pencils. In the first a profit of $ 5 per unit and the second $ 3 per unit is obtained. Both products go through three processes: Mold, assemble and test. Pens require 4 minutes, 2 minutes and 3 minutes respectively in these processes. Pens require 1 minute, 2 and 2 minutes respectively. The time available for each process is respectively 1.200 minutes, 1.000 minutes 1.200 minutes. Instructions: 1. Quantify data Minutos Requeridos | Horas Disponibles | Procesos | Bolgrafos (x) | Lapiceros (y) Moldear | 4x | 1y | 1,200 | Ensamblar | 2x | 2y | 1,000 | Probar | 3x | 2y | 1,200 | 2. Set the objective function $5x + $3y = f (x, y) 3. Establish restrictions Minutos Requeridos | Horas Disponibles | Procesos | Bolgrafos (x) | Lapiceros (y) | | Moldear | 4x | 1y | f 1,200 | Ensamblar | 2x | 2y | f 1,000 | Test | 3x | 2y | f 1,200 | x,0 y,0 4. Evaluate restrictions Molding restriction: 4x +1y = 1,200 horas x=0 4(0) +1y = 1,200 horas y = 1,200 horas y=0 4x +1(0) = 1,200 horas x = 1,200/4 x = 300 horas Assemble restriction: 2x + 2y = 1,000 horas x=0 2(0) +2y = 1,000 horas y = 1,000 /2 y = 500 horas y=0 2x + 2(0) = 1,000 horas x = 1,000/2 x = 500 horas Test Restriction: 3x + 2y = 1,200 horas x=0 3(0) +2y = 1,200 horas y = 1,200 /2 y = 600 horas y=0 3x + 2(0) = 1,200 horas x = 1,200/3 x = 400 horas 5. Make the restrictions graph 6. Evaluar los puntos extremos de la grfica $5x + $3y = f (x, y) Punto A A (0,0) $5(0) + $3(0) = f (x, y) $0 = f (x, y) Punto B B (0,500) $5(0) + $3(500) = f (x, y) $1,500 = f (x, y) Punto C 1) Ensamblar: 2x + 2y = 1,000 horas 2) Probar: 3x + 2y = 1,200 horas 1) - 2) 2x + 2y = 1,000 horas -3x - 2y = -1,200 horas x = 200 horas x=200horas 2(200) + 2y = 1,000 horas 400 + 2y = 1,000 horas 2y = 600 horas y = 300 horas C (200,300) $5(200) + $3(300) = f (x, y) $1,000 + $900 = f (x, y) $1,900 = f (x, y) Punto D 1) Moldear: 4x +1y = 1,200 horas 2) Probar: 3x + 2y = 1,200 horas 2x 1) - 2): 8x + 2y = 2,400 horas -3x - 2y = -1,200 horas 5x = 1,200 horas x = 240 horas x=240 4(240) +1y =1200 horas y= 1,200 - 960 y= 240 horas D (240,340) $5(240) + $3(340) = f (x, y) $1,200 + $1,020 = f (x, y) $2,220 = f (x, y) Punto E B (300,0) $5(300) + $3(0) = f (x, y) $1,500 = f (x, y) 7. Determine at what point is the optimal combination of production and what is this. (Optimal combination absence indicate at which point the maximum gain is obtained) | production | | point | pens | pens | gain | A|0|0|$0| B | 0 | 500 | $ 1.500 | C | 200 | 300 | $ 1.900 | D | 240 | 240 | $ 2.220 | E | 300 | 0 | $ 1.500 | The optimum combination is at point D where 240 pens and 240 pens produced by a gain of $ 2.220. 8. Indicate which is the value of the maximum gain and at what point is. The value of the maximum gain is $ 2.220 and is at point D