Expert: Paul Klarreich - 3/19/2010

Question
A store csn sell 500 pens, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price, it can sell 20 more pens. For what price should the pens be sold for maximum profit?
I tried to use this equation- p(x) = (500+20x)(100-x)- 35x
as revenue function would be (500+20x)(100-x)and production of pens is 35x. I set its dervative to 0 to solve for x. however I did not get the right answer. the right answer is $0.80. I will appreciate your help in this question. Thanks

Answer
Questioner: Joseph
Country: Canada
Category: Calculus
Private: No
Subject: max/min problems
Question: A store csn sell 500 pens, which cost the store 35 cents each, for $1 each. For each penny the store lowers the price, it can sell 20 more pens. For what price should the pens be sold for maximum profit?
I tried to use this equation- p(x) = (500+20x)(100-x)- 35x
as revenue function would be (500+20x)(100-x)and production of pens is 35x. I set its dervative to 0 to solve for x. however I did not get the right answer. the right answer is $0.80. I will appreciate your help in this question. Thanks
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The trick is not to do too much at one time.

Give things names: ('x' is not a good one to choose.)

Let N = number of pens sold.
Let P = price charged.
let D = decrease in price = 100 - P
Let I = increase in sales = 20(100 - P)
let F = profit = Revenue - cost

Then:

N = 500 + 20(100 - P)
R (revenue) = PN
F = PN - 35N = (P - 35)(500 + 20(100 - P))

F = (P - 35)(500 + 2000 - 20P)

R = (P - 35)(2500 - 20P)

R = 2500P - 20P^2 + 700P - 87500

R = 3200P - 20P^2  - 87500

dR/dP = 3200 - 40P

Set that = 0:

3200 - 40P = 0

p = 80