The captain of a river boat cruise charges $36 per person, including lunch. The cruise averages 300 customers per day. The captain is considering increasing the price. A survey of customers indicates that for every $2 increase, there would be 10 fewer customers. What increase in price would maximize revenue?

You need to try to set it up yourself.

$36/person \, \times \, 300\, people = 10800

$38/person \, \times \, 290\, people = 11020

in general

(36 + 2n) \, \times \, (300-10n) = T (T = total)

where n is the number of $2 steps.

-20n^2+240n+10800=T

differentiate and set to zero to get maximum (or possibly a minimum) \frac {dT}{dn}=0

Hopefully you know calculus. If not, post back and we'll try to solve it some other way.

-40n+240=0

n=6

The price to charge is, therefore, $48/person, and you'll have 240 people on the boat. You will also save money on lunch :D

You can check using 4 and 7 to assure yourself that n=6 is a maximum (it is).