62000 spectators With the ticket price at $11 average attendance 27000 price dropped to $8 attendance rose to 31000 Assume that attendance is linearly related to ticket price

Question

A baseball team plays in a stadium that holds 62000 spectators. With the ticket price at $11 the average attendance has been 27000. When the price dropped to $8, the average attendance rose to 31000. Assume that attendance is linearly related to ticket price.

What ticket price would maximize revenue?

Answer 1

Attendance is 27,000 for $11 ticket, and attendance is 31,000 for $8 ticket.

If attendance is linearly related to ticket price, for a ($11-$8=) $3 drop in ticket price there is a (31,000-27,000=) 4,000 attendance rise, which means (4,000/$3~=) 1,333 attendance rise per dollar decrease.

So we can formulize their relation (A=attendance, P=price, C=constant)

P = -(3/4000)A + C

if we put the values for $11 ticket

11 = -(3/4,000)*27,000 + C

C = 11 + 81/4 = 31.25

so the A-P relation formula is

P = 31.25 - (3/4000)A

or

A = -(4000/3)*(P - 31.25)

The revenue(R) is

R = A*P = -(4000/3)*(P- 31.25)*P = -(16000/3)*(4P² - 125P)

we need max(R)

so we need to find the P value for which the derivative of R equals to zero

R' = -(16000/3)*(8P - 125) = 0 => P = 125/8 = 15.625

For maximum revenue, ticket price has to be $15.63