Question 1058503
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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: 
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<pre>
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.
</pre>

You can look into this link

<A HREF=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</A>


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 !