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January 31, 2015

January 31, 2015

Posted by **Jake** on Wednesday, March 7, 2012 at 7:47pm.

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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?

- Calculus -
**MathMate**, Wednesday, March 7, 2012 at 8:20pmLet

c=cost of fare

P(c)=number of passengers at cost c

then

P(c)=4000-(c-1.50)*100/0.25

=4000-400c+600

=4600-400c

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.

- Calculus -
**Jake**, Wednesday, March 7, 2012 at 8:27pmI keep getting the wrong answer :| I don't know what I have done wrong...

- Calculus -
**MathMate**, Wednesday, March 7, 2012 at 8:50pmWhat did you get, and how did you get your answer?

- Calculus -
**Melinda**, Wednesday, March 7, 2012 at 8:54pmI used this

R(c)=(5000)(4600-400c)

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 (100-0.25) and (100+0.25)

- Calculus -
**MathMate**, Wednesday, March 7, 2012 at 11:24pmIt is not exactly a linear function.

P(c)=4600-400c

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(4600-400c)=4600c-400c^2

R'(c)=4600-800c=0 =>

c=5.75, an exorbitant price for a ride.

The corresponding ridership is

P(5.75)=4600-400*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.

- Calculus -
**Anonymous**, Friday, February 15, 2013 at 7:30amA pizza shop ia having

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