Posted by Jake on .
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?
Wednesday, March 7, 2012 at 8:20pm
c=cost of fare
P(c)=number of passengers at cost c
Differentiate R with respect to c, equate the derivative to zero and solve for c.
(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.
Wednesday, March 7, 2012 at 8:27pm
I keep getting the wrong answer :| I don't know what I have done wrong...
Wednesday, March 7, 2012 at 8:50pm
What did you get, and how did you get your answer?
Wednesday, March 7, 2012 at 8:54pm
I used this
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 (100-0.25) and (100+0.25)
Wednesday, March 7, 2012 at 11:24pm
It is not exactly a linear function.
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.
c=5.75, an exorbitant price for a ride.
The corresponding ridership is
The corresponding revenue is
R(6)=13200 < 13225
R(5.5)=13200 < 13225
So R(5.75) is indeed the maximum.
Friday, February 15, 2013 at 7:30am
A pizza shop ia having