Calculus question help?

Question

A bookstore can obtain the best-seller 20,000 Leagues Under the Majors from the publisher at a cost of $6 per book. The store has been offering the book at a price of $30 per copy and has been selling 160 copies per month at this price. The bookstore is planning to lower its price to stimulate sales and estimates that for each $2 reduction in the price, 20 more books will be sold per month. At what price should the bookstore sell the book to generate the greatest possible profit?

Answer 1

The hardest part of this problem is establishing the equation, and that is not super hard if you look at it piecewise.  Let's say profit is Y and the number of reductions is X.

 

y = (160+20x)(24-2x)  This means that the first instance (160 copies at a profit of $24 per book) is x=0

 

This can then be foiled and when you take the derivative you get dy/dx=-80x+160.  To get the maximum you need to set this equal to 0.  This will give you a final answer of x=2.

 

This means the highest profit will be from selling 200 books at $26

Answer 2

First, consider the relationship between price and sales which is linear. Let s = sales and p = price.

The equation goes through the points (30, 160) and (28, 180)

The slope is (160-180)/(30-28) = 10

The equation then is (s-180)/(p-28) = -10

s=460-10p    You should verify this!

 

Next find the revenue = p*s = 460p - 10p² 

 

The cost of the books sold is 6s = 6(460-10p) = 2760-60p

 

The profit = rev - cost = -10p² + 520p -2760

 

Since this is a calculus question:

The maximum profit will happen where the derivative = 0 i.e.  where -20p + 520 = 0

So the price to maximize profit is p = 26