A transit company carries about 80 000 riders per day for a fare of $1.25. To obtain more revenue, the management plans to increase the fare. It estimates that for every 5 cent increases in fare, it will lose 1000 riders. What fare would result in the greatest revenue and what would the maximum revenue be?

Question

anonymous · Answer

Let x be the times the company increases its far price. 
So lets say they increase by 15 cents than x=3 

What is the revenue of the company?
The number of riders*fare price. 
Currently it is: 80 000*1.25

After increasing the price by x*0.05 cents:
The riders will decrease by 1000*x and the fare will increase by x*0.05
Thus the new revenue is: 

(80 000-1000x)*(1.25+x0.05)      Expanding the bracket
100 000+4000x-1250x-50x^2
-50x^2 +2750x + 100 000       
Take out the common factor (-50 so we will have x^2)
-50*(x^2 -55x - 2000)

As we are looking for max revenue depending on x we can ignore 50

Max of:  - (x^2 -55x - 2000)
That is the Min of: x^2 -55x - 2000

Hint
(x-27.5)^2=x^2-55x^2 +756.25

x^2 -55x - 2000=(x-27.5)^2 -756.25 -2000

(x-27.5)^2 - 2756.25

Has minumum at x-27.5=0
x=27.5


Lets check 27.5*0.05=1.375 new price is: 1.25+1.375=2.625
80000-27.5*1000= 52500
New revenue: 52500*2.625=137 812.5

anonymous · Answer

thanks! really appreciate your help:D