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Show that if a rectangle has its base on the x-axis and two of its vertices on the curve y = e^-x^2 , then the rectangle will have the largest possible area when the two vertices are at the points of inflection of the curve.

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Hou

  1. let (x,y) be the point of contact in quadrant I, then (-x,y) is the other vertex.
    base of rectange = 2x
    height of rectange = y = e^(-x^2)

    area = 2xy
    = 2x(e^(-x^2))
    d(area)/dx = 2x(e^(-x^2))(-2x) + 2e(-x^2)
    = 0 for a max of area
    2e^(-x^2)[-2x^2 + 1] = 0
    2e^(-x^2) = 0 ---> no solution or

    -2x^2 + 1 = 0
    x = ± 1/√2
    then y = e^(-1/2) = 1/√e
    max area occurs when vertices are (1/√2, 1/√e) and (-1/√2, 1/√e)

    I will leave it up to you to differentiate
    y = e^(-x^2) twice, set that equal to zero, and show that x = ±1/√2

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    Reiny

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