# help me please?

A baseball team plays in a stadium that holds 66000 spectators. With the ticket price at $10 the average attendance has been 29000. When the price dropped to $7, the average attendance rose to 33000. Assume that attendance is linearly related to ticket price.

What ticket price would maximize revenue? $

### 2 Answers

- AlanLv 78 months ago
since Attendance is linear per prices

A = mp + b where p = ticket price

since attendance = people , then the slope will (people/price)

slope = (33000-29000) / (7-10) = -4000/3

A = (-4000/3)p + b

plug in one point

(10, 33000)

33000 = (-4000/3)*10 + b

33000 + (40000/3) = b

b = 46333 1/3

A = (-4000/3)p + 46333 1/3

R = Ap

R = (-4000/3)p^2 + (46333 1/3)p

where A is limited between 0 to 66000

so check the limits

if p =0 , you make no money

if A=0 , (-4000/3)p = -46333 11/3)

p = (-46333 1/3) *(-3/4000) = 34.75

at limit of p = 34.75 you make no money

so check for critical points

R' = (-8000/3)p + 46333 1/3

0 = (-8000/3) p + 46333 1/3

(8000/3) p = 46333 1/3

p = 46333 1/3 * ( 3/8000) =

p =17.375

but , you can't sell a ticket for have cent

so try p = 17.37 or 17.38

R = (-4000/3) *17.37^2 + 46333 1/3 * (17.37) = 402520.8 0

R = (-4000/3) *17.38^2 + 46333 1/3 * (17.38) = 402520.8 0

So p = $17.37 or 17.38 will both result in maximum revenue