# 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...

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.

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### 3 Answers

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.**

I do not think it is the right way to approach this question :( So far I have worked this

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

`=4000-400c+600`

`=4600-400c`

I don't know what to do after this. Anyone would like to help me please, I have a test tomorrow and my teacher said questions like this will be on the test.

Ok, begin by calculating how much the bus system makes each day. 4000x$1.50=$6000

Now, think about how much the 100 fewer people would cost the company.

3900x$1.50=$5850, which is a 150 dollar decrease

And the company would make $.25 more per passenger so...

3900x$1.75=$6825

Then calculate the same thing for the other scenario.

4100x$1.50=$6150

and,

4100x$1.25=$5125

As you can see, the $.25 increase is better for the company in the long run. So the bus system should charge $1.75 for bus fare to make the largest possible revenue.