Question 27259
We're looking for a hypothetical increase (or decrease) in bus fair, based on maximizing profit. Since the problem doesn't presume any additional overhead per rider (eg, no matter what happens we won't be adding buses or needing more drivers or any other expense) our profit is purely based on two factor, riders and the fare. To find profit, we multiply the two together. So if 800 riders pay $2.25 each, we'll be making:
{{{800*2.25=1800}}}
So far no problem, but we need to be able to see what happens when we change the fair. Let's call the fare increase x. So the new fare will be:
{{{2.25+x}}} 
But as the problem says, changing the fair will change the number of riders. Everytime x changes by 25 cents we lose 40 customers. A change of a dollar we'll lose 160 customers (160x).
When you subtract our lost riders from a base of 800:
{{{800-160x}}} 
We'll now multiply the fare (2.25+x) times the riders (800-160x):
{{{(x+2.25)(-160x+800)}}}

Solving graphically:
{{{graph( 300, 200, -2, 8, -500, 2400, (2.25+x)(800-160x))}}}
This shows that we should expect that an optimal fare increase of a little less than $1.50 will gross us the most money, approximately $2000.

Algebraically we want to find the maxima for the graph.
{{{y=(x+2.25)(-160x+800)}}}Factor:
{{{y=-160(x+2.25)(x-5)}}}FOIL:
{{{y=160(x^2-2.75x-11.25)}}}Completing the square:
{{{y/160+11.25=x^2-2.75x+121/64}}}
{{{y/160+11.25+(121/64)=(x-22/16)^2}}}Multiply everything by 160:
{{{y+1800+302.5=160(x-22/16)^2}}}
{{{y+2102.5=160(x-22/16)^2}}}
So your optimal fare increase is 22/16, or $1.375 or $2.25+1.375=$3.625. At that rate we'll have 22/16*160 or 220 fewer riders, or 800-220=580 riders. So we'll make 580 x 3.625 = $2102.5 profit. Of course we've got a problem that we have a half a penny to work with, as you can't have a $1.375 fair increase, and it's probably not even practical to have a fair increase that isn't an even nickel. So maybe you'd argue for a fair increase of $1.40 or $1.35 instead. Hope that helps.