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March 30, 2017

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Find the dimensions of the rectangle with the most area that will fit above the x-axis and below the graph of y=e^-x^2.

  • Calculus - ,

    The curve looks sort of bell-shaped.
    Let the point of contact in quadrant I be (x,y)

    then the base of the rectangle is 2x and its height is y
    A = 2xy
    = 2x(e^(-x^2))
    d(A)/dx = (2x)(-2x)(e^(-x^2)) + 2(e^(-x^2))
    = 0 for a max/min of A

    skipping some steps
    e^(-x^2)) ( -2x + 1) = 0

    x = 1/2
    then y = e^(-1/4) = 1/e^(1/4)

    so the rectangle has a length of 2x or 1
    and a height of 1/e^(1/4)

  • Calculus - ,

    Can you show me the steps you skipped?
    Thanks

  • Calculus - ,

    from
    (2x)(-2x)(e^(-x^2)) + 2(e^(-x^2)) = 0
    -2e^(-x^2) ( -2x^2 + 1) = 0

    ahhh, just noticed an error from earlier, (that's what I get by skipping steps)

    -2e^(-x^2) = 0 ----> no solution
    or
    -2x^2 + 1 = 0
    x^2 = 1/2
    x = 1/√2
    then y = e^(-1/2) = 1/√e

    so the rectangle has a length of 2x or 2/√2
    and a height of 1/√e

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