Question 202797: theater owner charges $3.00 of admission, there is an average attendance of 100 people. for every .10 increase in the admission price , there is a loss of 2 customers from the average number.
Teacher is asking what admission price should be charged in order to maximize revenue?
what is the maximum revenue?
I set it up
R(x)= (3.00-.10x)(100-2x)
am I on the right track
not sure what to due next
thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! theater owner charges $3.00 of admission, there is an average attendance of 100 people. for every .10 increase in the admission price , there is a loss of 2 customers from the average number.
Teacher is asking what admission price should be charged in order to maximize revenue?
what is the maximum revenue?
I set it up
R(x)= (3.00-.10x)(100-2x)
am I on the right track
-----------------------------------
Revenue = unit price*attendance
R = (300-10x)(100-2x)
R = 30000 - 600x -1000x + 20x^2
R = 20x^2 - 1600x + 30000
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Maximum R occurs when x = -b/2a = 1600/(40) = 40
When x = 40 the ticket price is 300-40 = $2.60
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Cheers,
Stan H.
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