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    kitteneater's Avatar
    kitteneater Posts: 4, Reputation: 1
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    #1

    Sep 20, 2008, 04:48 PM
    Pre-calc area in parabola
    I really need help on this problem. I have no idea how to do this.:
    A rectangle is inscribed in the region enclosed by the graphs of F(x)=18-x^2 and G(x)= 2x^2- 9. It gives a picture too, but there are no points on it for the rectangle. The rectangle is vertical though, with the longest legs being parallel to the Y axis, and the shortest to the X axis.


    A. Find the height of the rectangle as a function of X.
    B. Find the Width of the rectangle in terms of X.
    c. Write the area of the rectangle as a function of X.

    I think this has to do with optimization. But how would you find the max area of a shape in parabolas if you didn't have the points of where the rectangle intersected the parabola? I've already found the vertex points of the parabolas: ) (0, 19) and 0, -9. But does that even have anything to do with the problem? I'm so freakin' confused.

    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    Sep 20, 2008, 05:04 PM
    It's a shame you can't post the picture. Is the rectangle sitting on the x-axis? Or does it extend below the x-axis? If we let x be the distance on either side of the y-axis, then the area is 2xy. Sub in y=18-x^2.



    Now, the area is in terms of x.

    I am taking a leap here. It would help to see the picture.
    kitteneater's Avatar
    kitteneater Posts: 4, Reputation: 1
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    #3

    Sep 20, 2008, 05:39 PM
    It's not on the X axis. It sits just below. The rectangle seems like it's cut in half by the y axis too. http://s61.photobucket.com/albums/h6...nfusion012.jpg
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Sep 22, 2008, 05:46 AM

    No optimization needed - they simply want you to come up with the width and height for varoius width rectangles of half-width x. The height of the rectagle is merely F(x) - G(x). For example, for x =2: F(2)-G(2) = 18-2^2 - (2*2^2 - 9) = 14-(-1) = 15. And the width is simply 2x. So now you should be able to wite the formula for the area of the rectangle.
    luzcancino's Avatar
    luzcancino Posts: 1, Reputation: 1
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    #5

    Jan 14, 2009, 12:15 AM

    I need help on the parabola equations how to do a equation and what vertex , focus,and directrix are and where they go on the equation
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #6

    Jan 14, 2009, 07:38 AM
    Quote Originally Posted by luzcancino View Post
    i need help on the parabola equations how to do a equation and what vertex , focus,and directrix are and where they go on the equation

    Google. Lots of stuff out there.
    chexican's Avatar
    chexican Posts: 1, Reputation: 1
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    #7

    Sep 7, 2009, 09:33 PM
    Yes please I need help with the same thing
    STEP BY STEP

    Please! And thanks

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