Question

a) An airline will fill 100 seats of its aircraft at a fare of $200. For every $5 increase in the fare, the plane loses two passengers. For every decrease of $5, the company gains two passengers. What price maximizes revenue?

How Can I determine the number of passengers?

Answer 1

Fare of `$225` mximises revenue at `$20,250`. Number of passengers are `90`.

Explanation:

Let the fare br `f` and number of passengers be `p`. If `f` increases by `5`, `p` decreases by `2` and hence

`(Deltaf)/(Deltap)=-5/2` and relation between `f` and `p` is

`2f+5p=2xx200+5xx100=900` or `f=-5/2p+450`

and revenue `R` is given by `R=fxxp=-5/2p^2+450p`

and revenue maximises when `(dR)/(df)=-5p+450=0` or `p=90`

At this fare, number of passengers are `-5/2xx90+450=225`

and revenue is `20250`

graph{y+(5/2)x^2-450x=0 [0, 200, 18507, 20612]}