A basketball team plays in a stadium with a capacity of 55,000 spectators. With the ticket price at $10, the average attendance at recent games has been 27,000 spectators. Market research indicates that for every dollar the ticket price drops, attendance will increase by 3,000. Identify the function that maximizes revenue from ticket sales.

To maximize revenue from ticket sales, we need to find the ticket price that will result in the highest total revenue. Revenue is calculated as the product of the ticket price and the number of spectators. Let's denote: P as the ticket price (in dollars) A as the number of spectators (in thousands) R as the revenue (in dollars) We know that the stadium's capacity is 55,000 spectators, but the average attendance at recent games has been 27,000 spectators. This suggests that attendance (A) is affected by the ticket price (P) based on the market research. We can represent this relationship as follows: A(P) = 27 + 3(P - 10) Here, when the ticket price is $10, the attendance is 27,000 spectators. For every dollar the ticket price drops below $10, the attendance increases by 3,000 spectators. Now, we can express the revenue (R) as the product of the ticket price and attendance: R(P) = P * A(P) Substituting the expression for A(P) from above: R(P) = P * [27 + 3(P - 10)] R(P) = 27P + 3P^2 - 30P Now, simplify the equation: R(P) = 3P^2 - 3P - 30P R(P) = 3P^2 - 33P Therefore, the function that maximizes revenue from ticket sales is R(P) = 3P^2 - 33P