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.

Update:

What ticket price would maximize revenue? \$

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• 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

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• OK... So, what's your question?

• aurora3 weeks agoReport

What ticket price would maximize revenue? \$

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