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?

1 Answer
Aug 14, 2016

Answer:

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#.