Question 1058503: Brian runs a factory that makes Blu-ray players. Each S100 takes 6 ounces of plastic and 4 ounces of metal. Each D200 requires 3 ounces of plastic and 8 ounces of metal. The factory has 300 ounces of plastic, 608 ounces of metal available, with a maximum of 22 S100 that can be built each week. If each S100 generates $10 in profit, and each D200 generates $2, how many of each of the Blu-ray players should Brian have the factory make each week to make the most profit?
S100:
D200:
Best profit:
Answer by ikleyn(52878)   (Show Source):
You can put this solution on YOUR website! .
Brian runs a factory that makes Blu-ray players. Each S100 takes 6 ounces of plastic and 4 ounces of metal.
Each D200 requires 3 ounces of plastic and 8 ounces of metal. The factory has 300 ounces of plastic, 608 ounces of metal available,
with a maximum of 22 S100 that can be built each week.
If each S100 generates $10 in profit, and each D200 generates $2, how many of each of the Blu-ray players should Brian
have the factory make each week to make the most profit?
S100:
D200:
Best profit:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let x = # of S100 players to produce,
y = # of D200 players to produce.
Then the restrictions are
6x + 4y <= 300, (plastic)
3x + 8y <= 608. (metal)
There is also a restriction x <= 22 for for the number of S100 players.
Two other obvious restrictions are x >= 0 and y >= 0.
The objective function is z = 10x + 2y, which you must to maximize.
The setup is done.
The rest is just arithmetic, if you know what the LINEAR PROGRAMMING METHOD is.
You can look into this link
https://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.1058105.html
https://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.1058105.html
I solved there another problem, but you can still understand the idea of the LINEAR PROGRAMMING METHOD from there,
or refresh your knowledge.
Good luck !
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