SOLUTION: An amusement park usually charges $34 per ticket and averages 500 visitors per day. A study shows that if the park raises its
price by $1 per ticket, it will lose 125 visitors per
Question 1193232: An amusement park usually charges $34 per ticket and averages 500 visitors per day. A study shows that if the park raises its
price by $1 per ticket, it will lose 125 visitors per day. What price should the park charge per ticket, to maximize the
revenue?
Create an equation to solve this problem. Include appropriate “Let” statements to define your variables. Found 2 solutions by ankor@dixie-net.com, ikleyn:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! An amusement park usually charges $34 per ticket and averages 500 visitors per day.
A study shows that if the park raises its price by $1 per ticket, it will lose 125 visitors per day.
What price should the park charge per ticket, to maximize the revenue?
:
let x = no. of dollars per ticket raised
:
Revenue = ticket price * no. of visitors
R(x) = (34+x)(500-125x)
FOIL
R(x) = 17000 - 4250x + 500x - 125x^2
Arrange as a quadratic equation, R(x) = y
y = -125x^2 - 3750x + 17000
Find the Axis of symmetry using x = -b/(2a), where b=-3750, a=-125
:
It's obvious that his would not be a good business decision.
If you raised it 1 dollar and lost 125 visitors, the revenue would go down from 17000 to 13125 (35*375)
Let x be the increment of the ticket's price.
The revenue is the product of the price by the number of visitors
R(x) = (34+x)*(500-125x).
The roots of this quadratic function are x= -34 and x= = 4; so the optimum value of x is half-way between the roots
= = -15.
Thus the optimum price is 34 - 15 = 19 dollars. ANSWER
The revenue is then R(-15) = (34-15)*(500-125*(-15)) = 45125 dollars. ANSWER.
Compare it with the current revenue 34*500 = 17000 dollars.
Below is the plot for the revenue function R(x) = (34+x)(500-125x).
Notice that "x" in this plot is the of the price - - - it is NOT the price itself.
Plot R(x) = (34+x)(500-125x)
Thus decreasing of the ticket price results to increasing the number of visitors, and due to
increase the number of visitors the revenue grows, getting the maximum.
