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Math Help - Lower tick price, higher attendance. How do I relate them?

  1. #1
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    Lower tick price, higher attendance. How do I relate them?

    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. (Let x be the ticket price and R(x) be the revenue.)

    So far, I have the following (but can't put it all together):
    Revenue=attendance * price
    attendance = avg. attendance + 3000*(price reduction)

    t=10: tick price
    p: price reduction
    x: attendance
    R(x): Revenue in terms of attendance

    R(x)= xt=10x
    x=27000+3000p

    R(x)=270000+30000p
    Wrong letter ^ ^

    How do I relate t and p?
    3000*(t-p) gives a smaller number for a bigger p (wrong).
    3000*(p-t) gives negative numbers (for example p=1 is a $1 reduction means p-t=-9), but at least it gets bigger as p gets bigger.


    I could do these two myself if I could do (a).
    (b) Find the price that maximizes revenue from ticket sales.
    (c) What ticket price is so high that no revenue is generated?
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  2. #2
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    try this n= ax+b given (10,27000) and (9,30000)

    Then 27000 = 10a+b and 30000 = 9a+b

    Solve for a and b
    Last edited by pickslides; February 10th 2011 at 12:42 PM. Reason: Bad math..!
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  3. #3
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    Hello, MSUMathStdnt!

    A baseball team plays in a stadium that holds 55,000 spectators.
    With the ticket price at $10, the average attendance 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.
    . . Let \,x be the ticket price and \,R(x) be the revenue.

    Let \,x = ticket price.

    Note that: . x \:=\:10-r,\,\text{ where }r\text{ is the number of one-dollar reductions.}
    . . And we have: . r \:=\:10-x .[1]

    The survey indicates that if we reduce the ticket price by \,r dollars,
    . . attendance will increase by 3000r.


    \text{Revenue} \:=\:\text{Ticket price} \times \text{Attendance}

    . . . R(r) \;=\;(10-r)\cdot(27,\!000 + 3,\!000r)


    Substitute [1]: . R(x) \;=\;x\bigg[27,\!000 + 3,\!000(10-x)\bigg]

    Therefore: . R(x) \;=\;57,\!000x - 3,\!000x^2




    (b) Find the price that maximizes revenue from ticket sales.

    We have: . R(x) \;=\;57,\!000x - 3,\!000x^2


    Solve R'(x) = 0

    . . 57,\!000 - 6,\!000x \:=\:0 \quad\Rightarrow\quad x \:=\:9.5

    To maximize revenue, charge $9.50 per ticket.




    (c) What ticket price is so high that no revenue is generated?

    When is R(x) = 0\,?

    57,\!000x - 3,\!000x^2 \:=\:0 \quad\Rightarrow\quad 3000x(19-x) \:=\:0


    There are two equally silly scenarios:

    . . x\,=\,0 . . . . We give away free tickets!

    . . x \,=\,19 . . . . We charge $19 per ticket and no one comes!

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