Calculus
posted by Jake .
Could someone help me setup an equation for this equation?

A city bus system carries 4000 passengers a day throughout a large city. The cost to ride the bus is $1.50/person. The owner realizes that 100 fewer people would ride the bus for each $0.25 increase in fare, and 100 more people would ride for each $0.25 decrease in fare.
If the capacity of the bus system is 5000 passengers, what should the bus system charge to produce the largest possible revenue?

Let
c=cost of fare
P(c)=number of passengers at cost c
then
P(c)=4000(c1.50)*100/0.25
=4000400c+600
=4600400c
R(c)=c*P(c)
Differentiate R with respect to c, equate the derivative to zero and solve for c.
Note:
(1) c may or may not be a multiple of $0.25.
(2) The result may look ridiculous because the fixed and variable costs have not been accounted for. 
I keep getting the wrong answer : I don't know what I have done wrong...

What did you get, and how did you get your answer?

I used this
R(c)=(5000)(4600400c)
R'(c)=200,000
in the question it says:
The owner realizes that 100 fewer people would ride the bus for each $0.25 increase in fare, and 100 more people would ride for each $0.25 decrease in fare.
So wouldn't it be (1000.25) and (100+0.25) 
It is not exactly a linear function.
P(c)=4600400c
represents the number of riders.
There is nothing in the question that requires the ridership to be 5000 to get the maximum profit.
The revenue is the product of ridership, P(c), and the ticket price, c.
So revenue is indirectly a function of the ticket price, i.e.
R(c)=cP(c)=c(4600400c)=4600c400c^2
R'(c)=4600800c=0 =>
c=5.75, an exorbitant price for a ride.
The corresponding ridership is
P(5.75)=4600400*5.75=2300
The corresponding revenue is
R(5.75)=2300*5.75=13225
Check:
R(6)=13200 < 13225
R(5.5)=13200 < 13225
So R(5.75) is indeed the maximum. 
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