Thread: Revenue Word Problem

A local club is arranging a charter flight to Hawaii. The cost of the trip is $425 each for 75 passengers, with a refund of $5 per passenger for each passenger in excess of 75.

a. find the number of passengers that will maximize the revenue received from the flight.

b. find the maximum revenue

Originally Posted by shepherdm1270

A local club is arranging a charter flight to Hawaii. The cost of the trip is $425 each for 75 passengers, with a refund of $5 per passenger for each passenger in excess of 75.

a. find the number of passengers that will maximize the revenue received from the flight.

b. find the maximum revenue

There is something missing from this question.

All you have is the cost of the trip as a function of the number
of passengers (worded so badly that I can't tell you the cost for any
number of passengers over 75).

It looks to me as though the marginal cost per extra passenger is always
positive so there would be no maximum to the cost.

RonL

We did one in class, but my notes are kind of jumbled and I'm not sure how it was done. Here's what my notes looked like, see if this helps:

Profit: $5/seat if the seats are between 60-80, and it's decremented 5 cents per seat over 80.


Cost / # seats / profit
5.00 60 300
5.00 70 350
5.00 80 400

4.95 81 400.95
4.90 82 401.80


etc. etc.


SO price per seat = 5-(x-80)(.05)
R(x) = x(5-(x-80)(.05))
R9x) = 9x-.05x^2


----------------------------------------------------------

oh yeah, then:

R'(x) = 9-.10x
9-.10x = 0
9 = .10x
90 = x

ok i JUST worked on it some more, and i did the table and saw that 400 is the correct answer. then i worked on the formula and got this far:

425-(x-75)(5) = price per seat

R(x) = x (425 - (x - 75)(5))

So i need to simplify it, find the derivative, and solve for x...

the algebra is where i run into problems

help anyone? would be much appreciated. I'm freaking out. *sigh*

Originally Posted by shepherdm1270

ok i JUST worked on it some more, and i did the table and saw that 400 is the correct answer. then i worked on the formula and got this far:

425-(x-75)(5) = price per seat

R(x) = x (425 - (x - 75)(5))

So i need to simplify it, find the derivative, and solve for x...

the algebra is where i run into problems Mr F says: Fact of life. The further you go in your mathematical studies, the more it will be expected that you can do basic algebra. If you are learning calculus (which you appear to be doing) then it's absolutely expected that you have basic algebraic competencey.

Expand in side the brackets and simplify:

R(x) = x(800 - x)

=> R(x) = 800x - x^2.

The basic algebraic competency leading to this result is absolutely assumed in a course where calculus is being taught. You need to go back and thoroughly revise the basics.

In another thread Captain B has already given a thorough discourse on how to find the turning point of a quadratic.

this is the last math course I will ever be expected to take... I'm just trying to make it through. I'm 21 and haven't taken a math course since I was 17... needless to say, I'm a little rusty, and I'm not going to start over for one semesters worth of calculus... but thanks for the advice

well see I got :

x ( 425 - (x-75)(5))

x ( 425 - (5x - 375)

x ( 50 - 5x)

??

Originally Posted by mr fantastic

Expand in side the brackets and simplify:

R(x) = x(800 - x)

=> R(x) = 800x - x^2.

The basic algebraic competency leading to this result is absolutely assumed in a course where calculus is being taught. You need to go back and thoroughly revise the basics.

In another thread Captain B has already given a thorough discourse on how to find the turning point of a quadratic.

And I will take my own advice too!!

R(x) = x (425 - (x - 75)(5)) = x (425 - 5x - (-375)) = x (425 - 5x + 375) = x (800 - 5x).

NOT x (800 - x) as I first posted.

So R(x) = 800x - 5x^2.

R'(x) = 800 - 10x.

0 = 800 - 10x => 800 = 10x => x = 80.