Calculus problem need help

Question

(a) A cruise line offers a trip for  per passenger. If at least  passengers sign up, the price is reduced for all the passengers by  for every additional passenger (beyond ) who goes on the trip. The boat can accommodate  passengers. What number of passengers maximizes the cruise line’s total revenue? What price does each passenger pay then?

The number of passengers that maximizes the cruise line's total revenue is		.
The price the passengers pay is	.

(b) The cost to the cruise line for  passengers is . What is the maximum profit that the cruise line can make on one trip? How many passengers must sign up for the maximum to be reached and what price will each pay?

The maximum profit that the cruise line can make on one trip is 

Answer 1

a) P = 1700 - 10(n - 100)

P = 2700 - 10n

R = nP = n(2700 - 10n) = 2700n - 10n²

dR/dn = 0

2700 - 20n = 0

20n = 2700

n = 2700/20 = 135

P = 2700 - 10n

P = 2700 - 10(135)

P = 1350

b) Cost = 80000 + 500n

Let Pr be Profit

Pr = R - C = 2700n - 10n² - (80000 + 500n) = - 10n² + 2200n - 80000

d Pr/dn = -20n + 2200

20n = 2200

n = 110

They will pay 2700 - 10n = 2700 - 10(110) = 2700 - 1100 = 1600

Pr = - 10n² + 2200n - 80000 = -10(110²) + 2200(110) - 80000 = -121000 + 242000 - 80000 = 41000