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Math Help - Optimization Problem

  1. #1
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    Optimization Problem

    6. The base of a rectangle lies along the x axis, and the upper two vertices are on the curve defined by y = kx - x^2. Determine the dimensions of the rectangle with the maximum area.

    I started off with setting up the formula A = 2lw = 2xy and the constraint y = kx - x^2. After subbing y = kx - x^2 into A, I found the derivative of A and solved for x, which turned out to be 2/3k. I then subbed x into y = kx - x^2, which gave me a y value of 2k^2/9, and subsequently I found the maximum area to be 8k^3/27.

    The problem is that after I double checked my diagram, I found that it was inconsistent with the actual equation. Namely, it did not account for the fact that when x = 0, y = 0, and thus it had a large gap in the center.

    So my question is this: Did I set up my equations correctly? Is my parabola supposed to look something like http://jwilson.coe.uga.edu/emt668/EM...s/image007.jpg or something like http://xsquared.wikispaces.com/file/...l_parabola.jpg (but moved to the right a bit more)?
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  2. #2
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    Quote Originally Posted by RogueDemon View Post
    6. The base of a rectangle lies along the x axis, and the upper two vertices are on the curve defined by y = kx - x^2. Determine the dimensions of the rectangle with the maximum area.

    I started off with setting up the formula A = 2lw = 2xy and the constraint y = kx - x^2. After subbing y = kx - x^2 into A, I found the derivative of A and solved for x, which turned out to be 2/3k. I then subbed x into y = kx - x^2, which gave me a y value of 2k^2/9, and subsequently I found the maximum area to be 8k^3/27.

    The problem is that after I double checked my diagram, I found that it was inconsistent with the actual equation. Namely, it did not account for the fact that when x = 0, y = 0, and thus it had a large gap in the center.

    So my question is this: Did I set up my equations correctly? Is my parabola supposed to look something like http://jwilson.coe.uga.edu/emt668/EM...s/image007.jpg or something like http://xsquared.wikispaces.com/file/...l_parabola.jpg (but moved to the right a bit more)?
    w/o loss of generality, let k > 0

    y = kx - x^2 is a parabola that opens downward w/ x-intercepts at x = 0 and x = k

    base of inscribed rectangle = k - 2x , 0 < x < \dfrac{k}{2}

    height of inscribed rectangle = kx - x^2

    A = (k-2x)(kx-x^2)

    A = 2x^3 - 3kx^2 + k^2 x

    \dfrac{dA}{dx} = 6x^2 - 6kx + k^2 = 0

    x = \dfrac{k(3 - \sqrt{3})}{6}

    max area = \dfrac{k^3 \sqrt{3}}{18}
    Attached Thumbnails Attached Thumbnails Optimization Problem-parabolarectangle.jpg  
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  3. #3
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    Exactly what I needed! Thanks.
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