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The hypotenuse of a right triangle has one end at the origin and one end on the curve y+x^2e^(-3x), with x > or = 0. One of the other two sides is on the x-asis, the other side is parallel to the y-axis. Find the maximum area of such a triangle. At what x-value does it occur?

  • math/calc -

    Your curve
    y+x^2 e^(-3x)
    what is that equal to ? A curve has to follow an equation.
    If you mean the curve
    y=x^2 e^(-ex)

    then the area is yx/2 which is equal to
    1/2 x^3 e^(-3x). Right?

    dArea/dx= 3x^2e^(-3x)-3x^3e^(-3x)=0
    solve for x.
    x=1 check that I did it in my head. Mornings is not a great time for me to try that.

  • math/calc -

    You appear to have a typo,
    I will assume your curve is y = x^2e^(-3x) or y = x^2/e^(3x)

    let the point of contact of the hypotenuse be (x,y) on the curve.
    Then the right angle will be at (x,0)
    and the
    Area = xy/2
    = (1/2)(x)x^2e^(-3x)

    2A = (x^3)(e^(-3x))
    2dA/dx = x^3(-3e^(-3x)) + 3x^2(e^(-3x))
    = -3x^2(e^(-3x))[x - 1}
    = 0 for a max of A
    3x^2 = 0 --->x=0, little sense, since no triangle
    e^(-3x) = 0 ---> no solution
    x = 1, yeahh!

    if x = 1 then
    A = (1/2)1^3(e^-3) = .0249

    ( I tested for the area with x = .99 and x = 1.01 and they were both smaller than .0249 by a "smidgeon")

  • math/calc -

    thank you and yes i had atypo i apologized it was y = x^2e^(-3x

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