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Calculus BC: Rectangle inside an ellipse

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data
    What is thea are of the largest rectangle that can be inscribed in the ellipse 4x^2 +9y^2 = 36
    A) 6 rad 2 B) 12 C)24 D) 24 rad 2 E) 36


    2. Relevant equations
    Must be done using optimization and first derivitive


    3. The attempt at a solution
    I know I have to use A= 2x2y
    But i don't know how to remove one of the variables ..
     
  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    How about writing it as 4xy?

    Well, what equation do x and y satisfy if the corners of the rectangle are all points on the ellipse?
     
  4. Feb 18, 2009 #3
    x = radical ((36-9y^2)/4)
    but i'm not sure how that helps me.
    even when i plug that back in and take the derivitve, my answer isn't one of the multiple choice answers.

    thank you for the fast reply though :)
     
  5. Feb 18, 2009 #4

    Tom Mattson

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    It helps you because it eliminates one of the variables.

    Taking the derivative is only one step.

    Please post what you've done so far. That's the best way to see if you're making a mistake.
     
  6. Feb 18, 2009 #5
    A= 4xy
    A= 4 y (radical ((36-9y^2)/4))
    A'(y) = 6 radical 16-x^2 - (6x^2/radical 16-x^2) = 0
    when y = 2Rad2

    plug that back in : x = (radical ((36-9y^2)/4)) = 3rad2

    4(2rad2)(3rad2) = 48

    =/
     
  7. Feb 18, 2009 #6

    Tom Mattson

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    OK

    Where does [itex]\sqrt{16-x^2}[/itex] come from?
     
  8. Feb 18, 2009 #7
    the derivative of A(x)?
    it might be rad 4-x^2 instead... but regardless my answer still isn't there.
     
  9. Feb 18, 2009 #8

    Tom Mattson

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    Don't you think it makes a difference?? :eek:

    Yes it is. You need to correct that error and proceed from there. Please try it and if you get stuck, post your work. That's the only way I can spot a mistake.
     
  10. Feb 18, 2009 #9
    Tom, can you just please tell me if the answer is 12.

    I've done this problem at least 7 times, possibly 8.
    I have quite a few pages of work (with crossouts of course)

    I got y to be rad 2 and x to be rad 9/2

    so A = 4 xy
    A= 12?
     
  11. Feb 18, 2009 #10

    Tom Mattson

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    12 is what I got.
     
  12. Feb 18, 2009 #11
    yay, thank you so much =]

    would you like to give me a hint on another problem?

    If f is the continuous strictly increasing function on the interval a is less than or equal to x and x is less than or equal to b, which of the following must be true.

    I. integral from a to b f(x) dx < f(b) (b-a)
    II. integral from a to b f(x) dx > f(a) (b-a)
    III. integral from a to b f(x) dx = f(c) (b-a) for some number of c such that a<c<b

    I have no idea where to start and any hint would be wonderful.
     
  13. Feb 18, 2009 #12

    Tom Mattson

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    This problem is meant to direct your attention to the Mean Value Theorem for integrals.
     
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