# Optimization using differentiation

1. Oct 31, 2010

### A_Munk3y

1. The problem statement, all variables and given/known data

The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $600 per person per day if exactly 20 people sign up for the cruise. However, if more than 20 people sign up (up to the maximum capacity of 90) for the cruise, then each fare is reduced by$4 for each additional passenger.

(a) Assuming at least 20 people sign up for the cruise, determine how many passengers will result in the maximum revenue for the owner of the yacht.
passengers

(b) What is the maximum revenue?

(c) What would be the fare/passenger in this case? (Give your answer correct to the nearest dollar.)

3. The attempt at a solution
No idea, could you just help me out on how to even start this problem? Cause i am clueless...

i did it this way, but did not really use optimization i think.
N= # of passengers
600-4N= fare
Revenue= 600N-4N2
max at -> derivative of 0=600N-4N2
0=600-8N
N=600/8
N=75 passengers

600-4(75)= 300\$ a passenger

Last edited: Oct 31, 2010
2. Oct 31, 2010

### HallsofIvy

In order for there to be a problem, there must be a question! I see no question here. Was it to find the number of people who must sign up to maximize income?

Since the fare stays constant for up to 20 people, the more people you have the more income- so you can start with 20 people. If there are x people, with $x\ge 20$, then there are x- 20 people "more than 20" so the fare is reduced by 4(x- 20) dollars. That is, the fare for each of x people is 600- 4(x- 20) dollars per person. What is the total revenue? What is the maximum value for revenue?
(You don't really need Calculus for this. the total revenue function is quadratic and you can find the maximum value by completing the square.)

3. Oct 31, 2010

### A_Munk3y

Yea sorry, i forgot to put the 2nd part of the question.
I updated the question and what i did. She wants us to use optimization for this so she won't accept normal algebra answers :(