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Possible Maximum area

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A rectangle has its base on the x-axis and its vertices on the positive portion of the parabola
    $$ y=2-3x^2 $$
    What is the maximum possible area of this rectangle?
    A. (8/27)*181/2 B.(2/9)*181/2 C. (4/15)*301/2 D.(2/15)*301/2 E.(1/3)*121/2
    2. Relevant equations


    3. The attempt at a solution
    The definite integral (or the area underneath the curve) from 2/3 to-2/3, which are the x-intercepts, is 56/27 .Since the rectangle is inside the parabola, I just need to find the number that is the closest. It is choice C, but the answer turns out to be A.
     
    Last edited: Jul 11, 2013
  2. jcsd
  3. Jul 11, 2013 #2

    SteamKing

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    The intercepts of y = 2 - 3x^2 are NOT 2/3 and -2/3. Maybe this is your problem?
     
  4. Jul 11, 2013 #3
    If you draw a rectangle with base length 2x with the conditions you gave, the area of the rectangle is [itex]4x - 6x^3[/itex]. This function has derivative [itex]4 - 18x^2[/itex], which becomes zero at...

    Can you continue the solution?
     
  5. Jul 11, 2013 #4

    HallsofIvy

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    If the upper corners of the rectangle are at (x, y) and (-x, y) then the area of the rectangle is 2xy.

    Since [itex]y= 2- 3x^2[/itex], that area is [tex]2x(2- 3x^2)= 4x- 6x^3[/tex].
     
  6. Jul 11, 2013 #5
    So how do you guys use that expression to find its area? Do I derive it or integrate it?
     
  7. Jul 11, 2013 #6

    Mentallic

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    You now have a function

    [tex]A=4x-6x^3[/tex]

    Where A stands for area. For different values of x, you'll get a different value of A, correct? You want to find the maximum area the rectangle can be, hence, you want to find the maximum of A, so how do you do that?
     
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