A manufacture sells a certain article to dealers at a rate of $20 each , if less than 50 are ordered. If 50 or more are ordered (up to 600) the price per article is reduced at a rate of 2 cents times the number ordered. What size order will maximize revenue?
2006-11-26 07:27:41 · 2 answers · asked by Blesson 2 in Science & Mathematics ➔ Mathematics
For orders under 50,
f(x) = 20x
f'(x) = 20, which can never be 0, so there's no advantage here.
For orders over 50, (x here means number MORE than 50)
f(x) = (20 - .02x)x = 20x - .02x²
f'(x) = 20 - .04x = 0
Solving for x, .04x = 20, so x = 500
So the best value would be to order 550 articles.
There's some ambiguity here, though. The question doesn't say whether or not the first 50 articles will be discounted once you buy more than 50. I assumed that they wouldn't. If they would, then the answer would be just 500.
2006-11-26 07:35:14 · answer #1 · answered by Jim Burnell 6 · 0⤊ 0⤋
revenue between 50 and 600 is: x(20-0.02x)
take derivative and set it equal to zero:
20-0.04x=0
x=20/0.04=500
this assumes that you cannot buy more than 600
2006-11-26 07:32:50 · answer #2 · answered by Anonymous · 0⤊ 1⤋