SOLUTION: A bus company carries about 40 000 riders per day for a fee of $1.00. A survey indicates that if the fare is decreased , the number of riders will increase by 2500 for every 5 cent
Algebra ->
Human-and-algebraic-language
-> SOLUTION: A bus company carries about 40 000 riders per day for a fee of $1.00. A survey indicates that if the fare is decreased , the number of riders will increase by 2500 for every 5 cent
Log On
Question 784955: A bus company carries about 40 000 riders per day for a fee of $1.00. A survey indicates that if the fare is decreased , the number of riders will increase by 2500 for every 5 cents decrease. What fare will result in the greatest reverse?
Can you please help me out and can you show all the steps it would really help me understand. Thanks so much in advance:) Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A bus company carries about 40 000 riders per day for a fee of $1.00.
A survey indicates that if the fare is decreased, the number of riders will increase by 2500 for every 5 cents decrease.
What fare will result in the greatest reverse?
;
I'm not sure what "reverse" means here. Perhaps you mean "revenue".
:
Let x = number of 5 cent decreases and no. of 2500 passenger increases
:
Find the fare that gives maximum revenue
Revenue = fare * no. of passengers
R(x) = (1-.05x)(40000+2500x)
FOIL
R(x) = 40000 + 2500x - 2000x - 125x^2
Write as a quadratic equation
y = -125x^2 + 500x + 40000
Max y occurs at the axis of symmetry, find that using x = -b/(2a)
x =
x =
x = +2 fare decreases $.90 fare
:
That would give a increase in passengers of 5000
Max revenue: .90 *(40000+5000) = $40,500 max revenue
:
:
You can prove this to yourself, find the revenue using $.85 and $.95
