SOLUTION: A theater has 300 suscribers who each pay $400 for a season ticket. A theater wants to raise the ticket by $20. For each $20 increase tehy loose 10 suscribers. At what price can th

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A theater has 300 suscribers who each pay $400 for a season ticket. A theater wants to raise the ticket by $20. For each $20 increase tehy loose 10 suscribers. At what price can th      Log On


   

Question 316623: A theater has 300 suscribers who each pay $400 for a season ticket. A theater wants to raise the ticket by $20. For each $20 increase tehy loose 10 suscribers. At what price can they maxamize the revenue?
thanks a bunch!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
theater has 300 suscribers who each pay $400 for a season ticket.
A theater wants to raise the ticket by $20.
For each $20 increase they lose 10 subscribers.
At what price can they maximize the revenue?
:
Let x = no. $20 raises in price, and no. of 10 subscriber losses
:
Revenue = no. of subscribers * ticket price
r = (300-10x) * (400+20x)
FOIL
r = 120000 + 6000x - 4000x - 200x^2
A quadratic equation
r(x) = -200x^2 + 2000x + 120000
maximum occurs at the axis of symmetry, find that x = -b/(2a)
in this equation a = -200, b=2000
x = %28-2000%29%2F%282%2A-200%29
x = +5
:
Max revenue occurs when ticket price raise is $20 * 5 or $100 to $500
You will lose 5*10 = 50 subscribers or 250
so the revenue will be:
500 * 250 = $125,000
:
You can also check this by finding the vertex of this equation
Substitute 5 for x and solve for r(x)