Posted by cheryl on .
A company installs different POS computer systems. POS system A requires two hours to configure and one hour for assemble. POS B requires three hours to configure and one hour to assemble. POS C requires two hours to configure and two hours to assemble. The company has up to 100 labor hours for configure and 800 labor hours for assemble each week. If the profit of POS systems A, B, and C is $700, $ 800, and $1000 respectively how many of each system should be installed each week to maximize profit? What is the maximum profit each week?

finite math 
Anonymous,
This question is strange...why do the POS systems take longer to configure than they do to assemble, yet there are more labor hours for assembly than configuration. The minimum configuration time is two hours and the minimum assembly time is one hour. Yet for two hours config and two hours of assembly (one hour over minimum assembly) you get ~$300 more in profit. Thus, POS C will be the most profitable because the configuration hours determine maximum profit.
50 (100 configuration hours/2 hours per unit) POS C systems can be installed for a profit of $50,000 (assembly hours used = 100 which is less than 800)
33 POS B systems can be installed for a profit of $26,400 (assembly hours used = 99 which is less than 800)
50 POS A systems can be installed for a profit of $35,000 (assembly hours used = 100 which is less than 800)