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anonymous

  • 5 years ago

A baseball team plays in a stadium that holds 52000 spectators. With the ticket price at $12 the average attendance has been 19000. When the price dropped to $9, the average attendance rose to 26000. Assume that attendance is linearly related to ticket price. What ticket price would maximize revenue?

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  1. anonymous
    • 5 years ago
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    P= price x= attendance

  2. anonymous
    • 5 years ago
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    x = kP +c for constants k and c

  3. anonymous
    • 5 years ago
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    when P=12, x= 19000

  4. anonymous
    • 5 years ago
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    when P= 9 , x= 26000

  5. anonymous
    • 5 years ago
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    now you can form two eqns, solve for the constants c and k

  6. anonymous
    • 5 years ago
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    19000 =12k +c 26000= 9k +c subtract the first from the second 7000= -3k --> k = -7000/3

  7. anonymous
    • 5 years ago
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    sub that into one of the other eqns say the first , so c= 19000-12k = 47000

  8. anonymous
    • 5 years ago
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    thats what ive done so far but nothing from there is making sense. I also have to describe what im doing in words so i prove i understand it.

  9. anonymous
    • 5 years ago
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    so x= (-7000/3) P +47000

  10. anonymous
    • 5 years ago
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    revenue = R = attendance times cost price

  11. anonymous
    • 5 years ago
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    so R = P [ (-7000/3)P +47000 ] differentiate with respect to P , we want to maximise R while varying P

  12. anonymous
    • 5 years ago
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    R = (-7000/3) P^2 + 47000P dR/dP = (-14000/3)P +47000 =0 for min/max

  13. anonymous
    • 5 years ago
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    P = 3

  14. anonymous
    • 5 years ago
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    now check, second derivative is d^2R/dP^2 = (-14000/3) which is less than zero , so we do get a maximum . so ticket price = $3 maximises revenue

  15. anonymous
    • 5 years ago
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    wit its wrong

  16. anonymous
    • 5 years ago
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    should be P = (-47000) / ( -14000/3 ) = $10.07 , that maximises revenue

  17. anonymous
    • 5 years ago
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    $10.07 is the right answer. I still however have no idea how you got it. We haven't went over any problems by differentiating. I have no idea what that is.

  18. anonymous
    • 5 years ago
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    well , this can be done by other methods , I am assuming you have done quadratic equations before , ie , you know how to find the vertex of a parabola?

  19. anonymous
    • 5 years ago
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    and you know that the vertex ( which is a maximum or a minimum ) occurs on the axis of symmetry , and axis of symmetry of y= ax^2 +bx +c is x=-b / (2a)

  20. anonymous
    • 5 years ago
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    yes that's what we're doing now

  21. anonymous
    • 5 years ago
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    how did I know ;)

  22. anonymous
    • 5 years ago
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    so if you go back to this eqn R = (-7000/3) P^2 + 47000P and find the axis of symmetry , which is P = -b/ (2a) , this will give you the x coordinate of the maximum point

  23. anonymous
    • 5 years ago
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    which is 10.07, the answer

  24. anonymous
    • 5 years ago
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    thank you!

  25. anonymous
    • 5 years ago
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    Ok i know how you got this answer but why is there no c in the ax^2+bx+c equation? and how did you get 2 P's in the equation R= (-7000/3)P^2 +47000P ?

  26. anonymous
    • 5 years ago
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    "why is there no c? " , there just isnt "where did the two Ps come from" , look at how revenue was calculated, revenue = price x number of people

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