Question
Could someone help me setup an equation for this equation?
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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?
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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?
Answers
MathMate
Let
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.
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.
Jake
I keep getting the wrong answer :| I don't know what I have done wrong...
MathMate
What did you get, and how did you get your answer?
Melinda
I 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)
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)
MathMate
It is not exactly a linear function.
P(c)=4600-400c
represents the number of riders.
There is nothing in the question that <i>requires</i> 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.
P(c)=4600-400c
represents the number of riders.
There is nothing in the question that <i>requires</i> 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.
Anonymous
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