Question 1169652: A promoter of a show realizes that , if the price per ticket is set at 650.00 , only 50 people will watch the show . He also estimates that for every 50.00 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 ( number of tickets sold x price per ticket ) is a maximum ? What is the maximum value ?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! if the price per ticket is set at 650.00 , only 50 people will watch the show .
He also estimates that for every 50.00 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 (number of tickets sold x price per ticket) is a maximum ?
What is the maximum value?
Revenue = price * no. of tickets sold
R(x) = (650-50x)*(50+10x)
FOIL
R(x) = 32500 + 6500x - 2500x - 500x^2
arrange as a quadratic function
R(x) = -500x^2 + 4000x + 32500
Simplify, divide by -500
y = x^2 - 8x - 65
Find the the axis of symmetry
x = 8/(2*1)
x = 4
therefore
(10(4)+50) = 90 ticket sold for max revenue
and
(650-(4*50)) = $450 is the price of the tickets then
and
90 * 450 = $40500 is max revenue
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