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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.

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