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Quadratic equation word problem

  1. Apr 14, 2014 #1
    1. Problem
    Barkley runs a canoe-rental business on a small river in Pennsylvania. Currently, the
    business charges $12 per canoe and they average 36 rentals a day. A study shows that
    for every $.50 increase in rental price, the business can expect to lose two rentals per
    day. Find the price that will maximize income.

    2. The attempt at a solution
    I found this problem on the internet. I can probably solve the problem manually, but I want to learn how to create a set of equation to describe this situation.
    R= revenue, x= rental rate and t= number of rentals

    R = x * t
    R= 432 in a normal state.

    I am not sure how to link a $0.50 change in x to a 2 unit change in t.

    How should I go about this?

    Thanks.
     
  2. jcsd
  3. Apr 14, 2014 #2

    Mentallic

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    If you increase the price by $0.50 then your revenue will now be

    [tex]R=(x+0.5)(t-2)[/tex]

    Increase it again by another $0.50 and you now have?

    What if you increase it n times?
     
  4. Apr 14, 2014 #3

    Ray Vickson

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    Science Advisor
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    The demand (number of rentals per day) is given by
    [tex] N = 36 - 4(p-12)[/tex]
    where ##p =## price ($) and ##N=## number rented per day. Note that when ##p = 12## we have ##N = 36##, as given in the problem. Note also that ##N## decreases by 4 when ##p## increases by 1 (that is, ##N## decreases by 2 when ##p## increases by 0.5).

    The income is ##I = p N## because we have ##N## rentals and receive ##p##($) for each rental. Thus the daily income (in $) is
    [tex] I = pN = p[36 - 4(p-12)] = 84 p - 4 p^2 [/tex]
     
  5. Apr 17, 2014 #4
    Thank you.
     
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