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:

3 answers

Oh my heavens I have done so many linear programming problems tonight!
s uses 6 plastic
d uses 3 plastic
so
6 s + 3 d </= 300

s uses 4 metal
d uses 8 metal
so
4 s + 8 d </= 608

s </=22

maximize 10 s + 2 d

I get 332 at (22,56)
using
http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=1e692c6f72587b2cbd3e7be018fd8960&title=Linear%20Programming%20Calculator&theme=blue

use x for s and y for d
add
x>=0
y>=0

Thank You So Much!!!!!!

You are welcome.