Sally,

Look at
http://mathcentral.uregina.ca/QQ/database/QQ.09.08/h/karen2.html

Are you sure you have the problem worded correctly?

Harley

Sally wrote back.

Harley, yes the question is worded as written by my teacher. The airplane e question is different since you can't fill the plane fuller than the number of seats in the plane. What is your suggestion?

Thanks,
Sally

Sally,

The way the problem is worded if the order is for d dresses and d is larger than 100 then the cost to the manufacturer is $60 × d and the price of each dress is $89.10 so the revenue is $89.10 × d. Thus the profit is $89.90 × d - $60 × d = $39.90 × d. This is an increasing linear function of d and the profit is maximized by making as large an order as possible.

I think the problem should be worded so that the company will drop the price per dress by .10 for each dress ordered over 100. So if you order 101 dresses you pay $89.90 per dress. If you order 102 dresses you pay $89.80 per dress. If you order 103 dresses you pay $89.70 per dress, and so on.

For this problem let the number of dresses ordered be 100 + x. The manufactures cost is $60 × (100 + x). What is the cost per dress? What is the revenue function and what is the profit function? Use the calculus you know to maximize the profit.

If you need more assistance write back,
Harley