SOLUTION: A new movies theater after several months of incrementing finds that for every 20 c increase in ticket price the number of tickets for each showing decrease by 10. Similarly for ev

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Question 337535: A new movies theater after several months of incrementing finds that for every 20 c increase in ticket price the number of tickets for each showing decrease by 10. Similarly for every 20 c decrease in price showings increase by 10. On average the theater sells 500 tickets at $ 8 each.
how much should they charge for a ticket to maximize the revenue.?
Also explain how ticket price is related to total revenue?
Thank you
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Revenue = f ( Ticket Price, Tickets Sold ) = Price*Sold
Tickets Sold = f ( Price ) = m*Price + b
Sold = -10/(0.2)*(Price) +b
500 = -10/(0.2)*(8) + b
900 = b
Ticket Sold = -10/(0.2)*Price + 900
Tickets Sold = -50*Price + 900
Revenue = Price * Sold = Price * (-50*Price + 900)
Revenue = -50*Price^2 + 900*Price
Take the derivative and set to 0:
dRev = -100*Price + 900
0 = -100*Price+900
Price = 9.00
This is a maximum because the second derivative is negative:
d^2Rev = -100