Optimizing Ticket Prices for Maximum Revenue

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SUMMARY

The discussion focuses on optimizing ticket prices for a baseball team to maximize revenue based on attendance data. The demand function derived is p(x) = 8 - (1/6000)(x - 24000), indicating a linear relationship between ticket price and attendance. To maximize revenue, the optimal ticket price is calculated at $6.00 when attendance reaches 36,000 spectators. This solution effectively demonstrates the application of calculus in revenue optimization.

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[SOLVED] Optimization problem

Homework Statement


A baseball team plays in the stadium that holds 60000 spectators. With the ticket price at 8 the average attendence has been 24000. When the price dropped to 7, the average attendence rose to 30,000.
a) find the demand function p(x), where x is the number of spectators (assume p(x) is linear p(x)=
b) How should the ticket price be set to maximize revenue?


Homework Equations





The Attempt at a Solution


a) I figured out that the demand function was: 8-(1/6000)(x-24000)
b) I tried to use the revenue formula:; by taking the derivative of the demand function which is: 12-(12,000x/6000^{}2)
and then I set that equal to zero and tried to solve for X, which is: 36,000
So, now I don't know how to use the above information to calculate the price of the ticket. any suggestions?
 
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So x = 36,000 is a critical number of the revenue function r(x) = xp(x) = 12x - x^2 / 6000. What does that tell you about r(x) at x = 36,000? What does that tell you about p(x)?
 
I plugged in the x=36,000 into the demand function and I got $6.00 as the answer, I verified it and it is correct.
Thanks, I appreciate your help.
 

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