Stadium revenue

Question

A baseball team plays in a stadium that holds 55,000 spectators. With the ticket price at $10, the average attendance at recent games has been 27,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.

a) find a function that models the revenue in terms of ticket price. 

B) find the price that maximize revenue from ticket sales. 

C) what ticket price is so high that no revenue is generated?

Answer 1

A = Attendance, in terms of y

R = Revenue

TP = Ticket Price, in terms of x = 10

i.) R(x) = A(y) * TP(x)

 R(10) = 27,000 * 10 = $270,000 -- HIGHEST REVENUE
 R(9) = R(x-1) = (27,000 + 3,000) * (10 - 1) = 30,000 * 9 = $270,000 -- HIGHEST REVENUE
 R(8) = R(x-2) = (30,000 + 3,000) * (9 - 1) = 33,000 * 8 = $264,000
 R(7) = R(x-3) = (33,000 + 3,000) * (8 - 1) = 36,000 * 7 = $252,000
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R(1) = R(x-9) = (51,000 + 3,000) * (2 - 1) = 54,000 * 1 = $54,000

 

ii.) So, if attendance INCREASES by 3,000 for every $1 DECREASE in TP, we'll also see a DECREASE in attendance of 3,000 for every $1 INCREASE in TP within the function, as follows:

R(11) = R(x+1) = (27,000 - 3,000) * (10 + 1) = 24,000 * 11 = $264,000

R(12) = R(x+2) = (24,000 - 3,000) * (11 + 1) = 21,000 * 12 = $252,000

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R(18) = R(x+8) = (6,000 - 3,000) * (17 + 1) = 3,000 * 18 = $54,000

R(19) = R(x+9) = (3,000 - 3,000) * (18 + 1) = 0 * 19 = $0 -- LOWEST REVENUE

 

So, the Revenue is maximized when the TP is either $9 OR $10, based on the market survey research.

 

iii.) Based on the research, when the ticket price is TP = $19 no revenue is generated.