SOLUTION: A movie theater estimates that for each .50 cent increase in ticket price, the number of tickets sold decreases by 20. The current ticket price is 10.50 and the average daily ticke

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: A movie theater estimates that for each .50 cent increase in ticket price, the number of tickets sold decreases by 20. The current ticket price is 10.50 and the average daily ticke      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 767869: A movie theater estimates that for each .50 cent increase in ticket price, the number of tickets sold decreases by 20. The current ticket price is 10.50 and the average daily ticket sales is 1200 tickets. Set up an equation to represent the daily revenue as a function of the number of tickets sold. What price should the theater charge for tickets in order to maximize daily revenue?
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A movie theater estimates that for each .50 cent increase in ticket price, the number of tickets sold decreases by 20. The current ticket price is 10.50 and the average daily ticket sales is 1200 tickets. Set up an equation to represent the daily revenue as a function of the number of tickets sold. What price should the theater charge for tickets in order to maximize daily revenue?
***
R=revenue
let x=number of 50 cent increases
R=(1200-20x)(10.50+.50x)=12600+390x-10x^2
R=-x^2+39x+1260
R=-(x^2-39x+380.25)+380.25+1260
R=-(x-19.5)^2+1640.25
Maximum daily revenue of $1640.25 occurs when price of tickets=$19.50