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Equal area divided by line

  1. Apr 28, 2016 #1
    1. The problem statement, all variables and given/known data

    qppac1.png

    In the picture above, line y = c intersects with parabola y = 6x-x^2 in the first quadrant.
    If the gray area below line y = c and the gray area above line y=c are equal, then value of c is ...
    A. 19/4
    B.21/4
    C.23/4
    D.25/4
    E. 27/4

    2. Relevant equations

    Area under parabola = 2/3* base * height
    Area above parabola = 1/3 * base* height
    Area = definite integral

    3. The attempt at a solution

    Parabola y=6x-x^2 has peak point (3,9)

    Suppose a is the x-intercept of the line and parabola.
    At x = a, c equals 6a-a^2
    So, the gray area below the line is 1/3 * a * (6a-a^2)

    The base for the upper gray parabola area is 2(3) - 2a = 6-2a
    So, the gray area above the line is 2/3 * (6-2a) * (9-(6a-a^2))

    I get a = 1.592
    and c = 6a-a^2 = 7.017

    However, I'm sure the correct answer is 27/4 (I've tested it using integral method)
    Please help...
     
  2. jcsd
  3. Apr 28, 2016 #2

    Ray Vickson

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    Your expression for the gray area below the line is incorrect.
     
    Last edited: Apr 28, 2016
  4. Apr 28, 2016 #3

    blue_leaf77

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    Translate the parabola downward till the line y=c coincides with the x axis and then integrate the translated parabola from x=0 to the right intersection with the x axis. You will find c by making this integral vanish.
     
  5. Apr 28, 2016 #4

    blue_leaf77

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    Alternatively you can work with the inverse function ##x=f^{-1}(y)##. I think this way can br easier.
     
  6. Apr 29, 2016 #5
    Why??
    I think it's correct

    Okay, so the translated parabola become

    y = -x^2+6x-c

    But, I don't know the roots of the polynomial since there is an unknown c.
    Even if I substitute it with ( -6 ± 36 - 4k ) / -2 , I see it'll be very messy and hard to solve

    The inverse function of the parabola is

    f^-1(x) = √(9-x) + 3

    But, what does the inverse have to do with the area ? I don't understand the method
     
  7. Apr 29, 2016 #6

    blue_leaf77

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    I have tried following the method suggested in post #4. I ended up having to solve cubic polynomial in ##c##. However using graphical method I did find 27/4 as one of the solutions. Probably there are other members who can propose a more elegant method.
     
  8. Apr 29, 2016 #7
    Okay..
    Thanks anyway
     
  9. Apr 29, 2016 #8

    ehild

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    Anyway, you can write the solutions x1 and x2 (x1<x2) of the quadratic equation in terms of c. ##\int _0 ^{x_1}(c-y)dx = \int _{x_1} ^{x_2}(y-c)dx ## means that ## \int _0 ^ {x_2} {(-x^2+6x-c)dx}=0##
    Do the integration, substitute the limits, and solve the equation for c (it will become a quadratic one), not that terrible.
     
  10. Apr 29, 2016 #9

    blue_leaf77

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    I haven't tried myself what I suggested in post #3 which seems to coincide to your proposed method, but is it really quadratic in ##c##?
    I mean first we need to solve for ##x_2##, probably by using the usual quadratic root formula, and this will sure contain ##c## under a square root. As for the integral itself, there will be ##x^3## arising from the quadratic term in ##y(x)##.
     
  11. Apr 29, 2016 #10

    ehild

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    Yes, you proposed the same method.
    It will be simplified to a quadratic, as all terms of the definite integral contain x2.
     
  12. Apr 29, 2016 #11

    Ray Vickson

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  13. Apr 29, 2016 #12

    Thanks.. I've got it
     
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