Problem:
A commuter train line charges $1.50 per ride and averages 6000 passengers per day. They are contemplating a fare increase to maximize revenue. Study shows that $.25 increase in price results in a loss of 1000 passengers. What should the new fare be in order to maximize revenue?
I came up with this revenue function in order to solve. My prof didn't go over this scenario very well, so I'm not sure if I'm doing it correctly. Please let me know where/if I went wrong.
Let x=amount of increase
R(x)=(6000-4000x)(x+1.50)
I put "4000x" due to the fact that the increase will be in increments of .25, so if they raise the rate once, they will lose 1000 passengers, twice...2000...etc. I then took attempted to simplify (could be wrong):
R(x)=9000-4000x^2
..and took the derivative
R'(x)=8000x
...set it to zero (to find critical points)
8000x=0
...which would make x=0
So I came to the conclusion that the current price per ticket yields the commuter train line the maximum revenue. No change would be necessary. Is this correct?
Thanks in advance.
Edit:
After looking at my revenue function again, I think it should be:
R(x)=(1.50+.25x)(6000-1000x)
However, I still come to the same conclusion about the ticket price (no change needed).
A commuter train line charges $1.50 per ride and averages 6000 passengers per day. They are contemplating a fare increase to maximize revenue. Study shows that $.25 increase in price results in a loss of 1000 passengers. What should the new fare be in order to maximize revenue?
I came up with this revenue function in order to solve. My prof didn't go over this scenario very well, so I'm not sure if I'm doing it correctly. Please let me know where/if I went wrong.
Let x=amount of increase
R(x)=(6000-4000x)(x+1.50)
I put "4000x" due to the fact that the increase will be in increments of .25, so if they raise the rate once, they will lose 1000 passengers, twice...2000...etc. I then took attempted to simplify (could be wrong):
R(x)=9000-4000x^2
..and took the derivative
R'(x)=8000x
...set it to zero (to find critical points)
8000x=0
...which would make x=0
So I came to the conclusion that the current price per ticket yields the commuter train line the maximum revenue. No change would be necessary. Is this correct?
Thanks in advance.
Edit:
After looking at my revenue function again, I think it should be:
R(x)=(1.50+.25x)(6000-1000x)
However, I still come to the same conclusion about the ticket price (no change needed).