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.

Update:

What ticket price would maximize revenue? $

2 Answers

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  • Alan
    Lv 7
    8 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

  • JetDoc
    Lv 7
    8 months ago

    OK... So, what's your question?

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