math
posted by Britney .
The owner of a luxury motor yacht that sails among the 4000 Greek islands charges $960/person/day if exactly 20 people sign up for the cruise. However, if more than 20 people sign up (up to the maximum capacity of 100) for the cruise, then each fare is reduced by $8 for each additional passenger.
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
What is the maximum revenue?
$ ?
What would be the fare/passenger in this case? (Round your answer to the nearest dollar.)
? dollars per passenger

current: 20 passengers at $960 each
let the number of addional passengers be x
cost per passenger = 960  8x
revenue (R) = (20+x)(9608x)
= 19200  160x + 960x 8 x^2
dR/dx = 160 + 960  16x = 0 for a max of R
16x = 800
x = 50
There should be an additional 50 or a total of 70 passengers
the cost per passenger would be 9608(50) or $560 per day
check:
try a number around 70 passengers
for 69 passengers
cost = 9608(49) = 568
R = 39192
for 70 passengers
cost = 9608(50) = 560
R = 39200  the highest
for 71 passengers
cost = 96051(8) = 552
R = 39192