Question 1184212: Diana runs a factory that makes DVD players. Each R80 takes 6 ounces of plastic and 2 ounces of metal. Each FS20 requires 2 ounces of plastic and 4 ounces of metal. The factory has 140 ounces of plastic, 200 ounces of metal available, with a maximum of 12 R80 that can be built each week. If each R80 generates $5 in profit, and each FS20 generates $7, how many of each of the DVD players should Diana have the factory make each week to make the most profit?
R80:
FS20:
Best profit:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! make a table such as the one below:
product name R80 FS20
number of units x y x >= 0, y >= 0, x <= 12
plastic 6 2 6x + 2y <= 140
metal 2 4 2x + 4y <= 200
profit 5 7 maximize 5x + 7y
objective function is profit = 5x + 7y
this is what you want to maximize.
constraint functions:
x >= 0
y >= 0
x <= 12
6x + 2y <= 140
2x + 4y <= 200
using the desmos.com calculator, you graph the opposite of the constraint functions and evaluate each corner point of the feasible region with the objective function.
the feasible region is the area of the graph that is not shaded.
here's the graph.
the corner points of the feasible region and the value of the objective function at those corner points are:
(0,50) = 5x + 7y becomes 5*0 + 7*50 = 350
(8,46) = 5x + 7y becomes 5*8 + 46*7 = 362 ***** maximum profit is here.
(12,34) = 5x + 7y becomes 5*12 + 7*34 = 298
(12,0) = 5x + 7y becomes 5*12 + 7*0 = 60
the maximum profit is at (8,46), where it is equal to 362.
(8,46) means the value of x is 8 and the value of y is 46.
the constraint functions must all be true at this point.
x <= 12 is true.
x >= 0, y >= 0 is true.
6x + 2y becomes 6*8 + 2*46 = 140 which is <= 140, so this is true.
2x + 4y becomes 2*8 + 4*46 = 200 which is <= 200, so this is true.
all constraints are true at the maximum profit point of (8,46).
this confirms the solution at (8,46) is valid.
|
|
|
