1. Apr 14, 2014

### musicgold

1. Problem
Barkley runs a canoe-rental business on a small river in Pennsylvania. Currently, the
business charges $12 per canoe and they average 36 rentals a day. A study shows that for every$.50 increase in rental price, the business can expect to lose two rentals per
day. Find the price that will maximize income.

2. The attempt at a solution
I found this problem on the internet. I can probably solve the problem manually, but I want to learn how to create a set of equation to describe this situation.
R= revenue, x= rental rate and t= number of rentals

R = x * t
R= 432 in a normal state.

I am not sure how to link a $0.50 change in x to a 2 unit change in t. How should I go about this? Thanks. 2. Apr 14, 2014 ### Mentallic If you increase the price by$0.50 then your revenue will now be

$$R=(x+0.5)(t-2)$$

Increase it again by another $0.50 and you now have? What if you increase it n times? 3. Apr 14, 2014 ### Ray Vickson The demand (number of rentals per day) is given by $$N = 36 - 4(p-12)$$ where $p =$ price ($) and $N=$ number rented per day. Note that when $p = 12$ we have $N = 36$, as given in the problem. Note also that $N$ decreases by 4 when $p$ increases by 1 (that is, $N$ decreases by 2 when $p$ increases by 0.5).

The income is $I = p N$ because we have $N$ rentals and receive $p$($) for each rental. Thus the daily income (in$) is
$$I = pN = p[36 - 4(p-12)] = 84 p - 4 p^2$$

4. Apr 17, 2014

Thank you.