SOLUTION: A bus company has 4000 passengers daily, each paying a fee of 2$. For each 0.15$ increase, the company estimates that it will lose 40 passengers. If the company needs to take in $1

Question 384489: A bus company has 4000 passengers daily, each paying a fee of 2$. For each 0.15$ increase, the company estimates that it will lose 40 passengers. If the company needs to take in $10 450 per day to stay in business, what fare should be charged?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A bus company has 4000 passengers daily, each paying a fee of 2$.
For each 0.15$ increase, the company estimates that it will lose 40 passengers.
If the company needs to take in $10 450 per day to stay in business, what fare should be charged?
:
Use the given information write a revenue equation
Let x = no. of .15 increases above the 2.00 fare,
and x = no. of 40 passenger decreases from 4000
let y = daily revenue
:
The equation:
Revenue = fare * no. of passengers
y = (2 + .15x) * (4000 - 40x)
FOIL
8000 - 80x + 600x - 6x^2 = y
:
-6x^2 + 520x + 8000 = y
:
"What fare increase will produce $10450 in daily revenue?+
y = 10450, find x
-6x^2 + 520x + 8000 = 10450
:
-6x^2 + 520x + 8000 - 10450 = 0
:
-6x^2 + 520x - 2450 = 0
Simplify, divide by -2
3x^2 - 260x + 1225 = 0
Factor
(3x - 245) (x - 5) = 0
The reasonable solution
x = 5 ea .15 raises in fare will produce $10,450 revenue
Fare = 2.00 + 5(.15)
Fare = $2.75, is the required fare
:
:
Check this using x = 5
5 ea 40 passenger decreases, from 4000 equal 3800 passengers
Rev = 2.75 * 3800
Rev = $10,450
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