If the capacity of the bus system is 5000 passengers, what should the bus system charge to produce the largest possible revenue in the following case: 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.

With a fare of \$1.5 there are 4000 passengers of the bus service.

A \$0.25 increase in fare decreases the number of passengers by 100 and a \$0.25 decrease in fare increases the number of passengers by 100.

For a change in the fare of x, the total revenue

R = (4000 - 100*x)(1.5 + x*0.25)

=> 6000 - 150x + 1000x - 25x^2

To maximize R, solve R' = 0

R' = -150 + 1000 - 50x

-150 + 1000 - 50x = 0

=> 50x = 850

=> x = 17

The price of ticket should be increased by \$4.25 to \$5.75

Revenue is maximized when the fare is increased to \$5.75.

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