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# Word Problems/Algebra

Question
A manufacturing company has an order for 290 boxes.  Using a machine, they can produce 30 boxes per hour at a cost of \$15 per hour.  Using two student interns, together they can make 25 boxes per hour at a cost of \$10 per hour.  The machine and team of interns can work no more than 8 hours, each.  What is the most economical (i.e., least expensive) combination of machine and intern work times to meet the order?

Create a system of inequalities and solve by graphing to get your answer.

Sorry about the delay, but I've just got back from vacation.

As I understand this problem, the two interns working together can make 25 boxes an hour at a total pay of \$10.  Since there's two of them, they can each make 12.5 boxes per hour for \$5 each.  That puts the cost of 1 box at 5/12.5 = 40c each box.

The machine makes 30 boxes per hour at a cost of \$15,
so the cost of each is 15/30 = 50c each box.

From this, it can be seen that the interns are cheaper.  Since they make 25 boxes / hour and there are 8 hours, that gives us 8*25 = 200 boxes.

That leaves 290 - 200 = 90 boxes for the machine.  Since the machine is 30 boxes per hour,
that is 90/30 = 3 hours.
Questioner's Rating
 Rating(1-10) Knowledgeability = 9 Clarity of Response = 10 Politeness = 10 Comment Thank you, very much!

Word Problems

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#### Scott A Wilson

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