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Math Help - Max square/rec area under the curve

  1. #1
    Junior Member mant1s's Avatar
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    Max square/rec area under the curve

    Hi everyone, I've been working on this optimization problem for awhile now. I guess I don't understand the steps. Here's my work.

    The Problem
    Code:
    A rectangle is inscribed with its base on the x-axis and 
    its upper corners on the parabola y = 4 - x^2.  What are 
    the dimensions of such a rectangle with the 
    greatest possible area?
    My work

    Code:
    Define variables...
    A = 2xy
    y = 4- x^2
    
    setup problem:
    
    A(x) = 2x ( 4 - x^2)
    
    Find d/dx:
    
    A'(x) = 8 - 6x^2
    
    Critical Points:
     0 = A'(x) = { -1.1547, 1.1547}
    
    ~~Then am I supposed to use this formula??? ----> 4x + 3y ?
    Thanks for your time

    -M
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mant1s View Post
    Hi everyone, I've been working on this optimization problem for awhile now. I guess I don't understand the steps. Here's my work.

    The Problem
    Code:
    A rectangle is inscribed with its base on the x-axis and 
    its upper corners on the parabola y = 4 - x^2.  What are 
    the dimensions of such a rectangle with the 
    greatest possible area?
    My work

    Code:
    Define variables...
    A = 2xy
    y = 4- x^2
     
    setup problem:
     
    A(x) = 2x ( 4 - x^2)
     
    Find d/dx:
     
    A'(x) = 8 - 6x^2
     
    Critical Points:
     0 = A'(x) = { -1.1547, 1.1547}
     
    ~~Then am I supposed to use this formula??? ----> 4x + 3y ?
    Thanks for your time

    -M
    You don't need to use the formula 4x + 3y at all.

    You are exactly right, the Area of the rectangle is given by

    A = 2x ( 4 - x^2)

     = 8x - 2x^3.


    So to find the maximum

    \frac{dA}{dx} = 8 - 6x^2

    0 = 8 - 6x^2

    6x^2 = 8

    x^2 = \frac{4}{3}

    x = \pm \sqrt{\frac{4}{3}}

    x = \pm \frac{2\sqrt{3}}{3}.


    The maximum area would be at the point x = \frac{2\sqrt{3}}{3}.


    So put it into the formula for area.

    A = 8x - 2x^3

    A = 8\left(\frac{2\sqrt{3}}{3}\right) - 2\left(\frac{2\sqrt{3}}{3}\right)^3
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  3. #3
    Junior Member mant1s's Avatar
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    Thank you so much!
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  4. #4
    Junior Member mant1s's Avatar
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    I got 6.1584 for the MAX

    How would you find the Length and Width of the rectangle after that?

    -M
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  5. #5
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    well, if you look at prove it's post, the value of x that gives you the maximum area is already there. when x is that value, what is y? whichever is the greater of the two values becomes the length, the other is the width.
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  6. #6
    Junior Member mant1s's Avatar
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    So am I looking to do an perimeter calculation with my MAX (X)? Using the 4x + 3y formula. Also how do I get y, just algebra?

    Thx!

    -M
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  7. #7
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    well, you see, how you find y is already in the question itself.

    y = 4 - x^2.

    [edit] From there you can find x through the formula for area, as HallsofIvy mentions below.[/edit]
    Last edited by compliant; July 24th 2009 at 08:22 AM.
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  8. #8
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    Notice that while the height of the rectangle is y, the base length is 2x, not just x. The original problem says nothing about the perimeter. I don't understand why you keep mentioning it.
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  9. #9
    Junior Member mant1s's Avatar
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    hey sweet!

    I got...

    y= 2.66667
    x= 2.3094
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