Question

A transit company charges $1.25 per ride with 10 000 passengers a day. The company wants to increase profit but realize that with each 10 cent augmentation in fare, they will lose 500 passengers. What should the company charge to maximize profit?

Answer 1

Profit would maximize at `1.25+0.375=1.625` per passenger at which maximum revenue would be `$13203.125`.

Explanation:

Let the prices be increased in steps of `10` cents. If prices are increased by `x` such steps, the price becomes `(1.25+x/10)` per ride.

But number of passengers will go down by `500x` and revenue will be

`(1.25+x/10)(10000-500x)` or

`(125+10x)(100-5x)` - dividing second binomial by `100` and multiplying first by `100`. This is equivalent to

`12500+1000x-625x-50x^2` or

`12500+375x-50x^2`

Now we should have an extrema where first derivative is zero and as it is `375-100x=0` or `x=3.75`.

Note that second derivative is `-100` and as such it is a maxima.

Hence, profit would maximize at `1.25+0.375=1.625` per passenger at which maximum revenue would be `1.625(10000-500×3.75)=13203.125`.