SOLUTION: A promoter of a show realizes that, if the price per ticket is set at $650, only 50 people will watch the show. He also estimates that for every $50 decrease in the price of the ti

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Question 1125210: A promoter of a show realizes that, if the price per ticket is set at $650, only 50 people will watch the show. He also estimates that for every $50 decrease in the price of the ticket, the number of people watching the show will increase by 10. What should be the price per ticket so that the revenue is a maximum? What is the maximum revenue?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
If n is number of increments of 50 dollars, then price becomes 650-50n.
If use x for counting each increment of $50 then x=50n; and number of tickets is 50+10n. 
PRICE       TICKETS
650-x       50+x/5

Revenue is y=%28650-x%29%2850%2Bx%2F5%29.
Maximum revenue is exactly in the middle of the two zeros of y.

Zeros are 650 and -250.
Exact Middle of the two zeros: 200 dollars decrease from $650. Price there for max revenue, $450, best ticket price.
Max Revenue: $40500.