Question 1049512: A bus company has 4000 passengers daily, each paying a fare of $2. For each $0.15 increase, the company estimates that it will lose 40 passengers per day. If the company needs to have a revenue of $10 450 per day to stay in business, what fare should be charged?
(Functions Math)
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 fare of $2.
For each $0.15 increase, the company estimates that it will lose 40 passengers per day.
If the company needs to have a revenue of $10,450 per day to stay in business, what fare should be charged?
:
Rev = no. of pass * fare
the equation for this situation
:
let x = the no. of .15 increases and = no. of 40 passenger decreases
(4000-40x)(2+.15x) = 10450
FOIL
8000 + 600x - 80x - 6x^2 = 10450
Combine like terms to form a quadratic equation
-6x^2 + 520x + 8000 - 10450 = 0
-6x^2 + 520x - 2450 = 0
solve this using the quadratic formula. a=-6; b=520; c=-2450
I got two solutions
x = 81
and
x = 5, this is the reasonable solution. 5 price increases will yield $10,450
:
:
See if that's true
5 price increases: 5(.15) + 2 = $2.75 is the fare
5 40 pass decreases: 4000 - 5(40) = 3800 passengers
Rev = 2.75 * 3800 = $10,450
|
|
|
