A soccer stadium holds 48000 spectators.

Question

A soccer stadium holds 48000 spectators. With a ticket price of $15, the average attendance has been 28000. When the price dropped to $12, the average attendance rose to 31000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?



1. A .Find a linear function of the form  that relates the attendance , in thousands of spectators, to the price  of a ticket. 
2. B. Use your answer to part a., along with the fact that , to form the revenue function. 
3. C. Find the number of spectators needed to maximize revenue.Revenue is maximized when  spectators attend. (Remember that  is in thousands of spectators.)
4. D. Find the ticket price that will maximize revenue.Revenue is maximized when tickets cost  . (Use a “$” symbol in your answer.)
5. E. What is the maximum revenue?The maximum revenue is  . (Use a “$” symbol in your answer.)

Answer 1

(15,28000)

(12,31000)

slope m = 3000/-3 = -1000

B = y - mx = 28000 - (-1000)(15)

= 28000 + 15000

= 43000

y = -1000 x + 43000 where x is the price and y is the demand

D = -1000*P + 43000 where P is the price and D is the demand

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The inverse is (D-43000)/(-1000) = P

The revenue function is (x-43000)(-1000) x = R(x) where x is the attendance

(-1000x)(x-43000) = R(x)

-1000x^2 + 43000x

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The max occurs at the AVERAGE of the solutions. The solutions are x=43000 and x=0

The average is 43000/2 = 21500

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From part 2, P = (21500-43000)/-1000 = $21.50

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the max revenue is then 21.50 * 21500 = $462250