a television manufacturer makes qled and oled televisions. the profit per unit is $125 for the qled…

a television manufacturer makes qled and oled televisions. the profit per unit is $125 for the qled televisions and $190 for the oled televisions. let x = the number of qled televisions manufactured in a month and let y = the number of oled televisions manufactured in a month. complete parts (a) through (e).\na. write the objective function that models the total monthly profit.\nz = 125x + 190y\n(do not include the $ symbol in your answer.)\nb. the manufacturer is bound by the following constraints. write an inequality that models each constraint.\nequipment in the factory allows for making at most 450 qled televisions in one month.\nx≤450\nequipment in the factory allows for making at most 200 oled televisions in one month.\ny≤200\nthe cost to the manufacturer per unit is $600 for the qled televisions and $900 for the oled televisions. total monthly costs cannot exceed $360,000.\n600x + 900y≤360000\nc. graph the system of inequalities in part (b). use only the first quadrant and its boundary, because x and y must both be nonnegative.\nuse the graphing tool to graph the system. graph the region that represents the correct solution only once.\nd. evaluate the objective function for total monthly profit at each of the five vertices of the graphed region. what are the resulting values?\n(simplify your answers. use a comma to separate answers as needed. do not use commas in the individual numbers.)

a television manufacturer makes qled and oled televisions. the profit per unit is $125 for the qled televisions and $190 for the oled televisions. let x = the number of qled televisions manufactured in a month and let y = the number of oled televisions manufactured in a month. complete parts (a) through (e).\na. write the objective function that models the total monthly profit.\nz = 125x + 190y\n(do not include the $ symbol in your answer.)\nb. the manufacturer is bound by the following constraints. write an inequality that models each constraint.\nequipment in the factory allows for making at most 450 qled televisions in one month.\nx≤450\nequipment in the factory allows for making at most 200 oled televisions in one month.\ny≤200\nthe cost to the manufacturer per unit is $600 for the qled televisions and $900 for the oled televisions. total monthly costs cannot exceed $360,000.\n600x + 900y≤360000\nc. graph the system of inequalities in part (b). use only the first quadrant and its boundary, because x and y must both be nonnegative.\nuse the graphing tool to graph the system. graph the region that represents the correct solution only once.\nd. evaluate the objective function for total monthly profit at each of the five vertices of the graphed region. what are the resulting values?\n(simplify your answers. use a comma to separate answers as needed. do not use commas in the individual numbers.)

Answer

Explanation:

Step1: Identify the vertices of the feasible - region

The constraints are (x\leq450), (y\leq200), (600x + 900y\leq360000) (or (2x + 3y\leq1200)) and (x\geq0), (y\geq0). The vertices are found by solving the systems of equations of the boundary - lines.

  1. Intersection of (x = 0) and (y = 0): The vertex is ((0,0)).
  2. Intersection of (x = 0) and (y=200): The vertex is ((0,200)).
  3. Intersection of (y = 0) and (x = 450): The vertex is ((450,0)).
  4. Intersection of (x = 0) and (2x+3y = 1200): Substitute (x = 0) into (2x+3y = 1200), we get (3y=1200), so (y = 400). But since (y\leq200), this intersection is not a vertex of the feasible - region.
  5. Intersection of (y = 0) and (2x+3y = 1200): Substitute (y = 0) into (2x+3y = 1200), we get (2x=1200), so (x = 600). But since (x\leq450), this intersection is not a vertex of the feasible - region.
  6. Intersection of (y = 200) and (2x+3y = 1200): Substitute (y = 200) into (2x+3y = 1200), we have (2x+3\times200=1200), (2x+600 = 1200), (2x=600), (x = 300). The vertex is ((300,200)).
  7. Intersection of (x = 450) and (2x+3y = 1200): Substitute (x = 450) into (2x+3y = 1200), we get (2\times450+3y = 1200), (900+3y = 1200), (3y = 300), (y = 100). The vertex is ((450,100)).

Step2: Evaluate the objective function (z = 125x+190y) at each vertex

  • At ((0,0)): (z=125\times0 + 190\times0=0).
  • At ((0,200)): (z=125\times0+190\times200 = 38000).
  • At ((450,0)): (z=125\times450+190\times0=56250).
  • At ((300,200)): (z=125\times300+190\times200=37500 + 38000=75500).
  • At ((450,100)): (z=125\times450+190\times100=56250+19000 = 75250).

Answer:

0,38000,56250,75500,75250