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Math Help - Optimization and Modeling

  1. #1
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    Optimization and Modeling

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


    All right, I know that I'll probably need A = (1/2)bh. I know I have to in some way find out where the curve intersects either the vertical side of the triangle or the hypotenuse to maximize the area? I think? But I don't know if i'm thinking the right way, and I don't know how to go about setting this up.
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  2. #2
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    Hello, Jacobpm64!

    It's easier than you think . . .


    The hypotenuse of a right triangle has one end at the origin
    and one end on the curve y = x^2e^{-3x}, with x \geq 0.
    One of the other two sides is on the x-axis, the other side is parallel to the y-axis.
    Find the maximum area of such a triangle. At what x-value does it occur?
    Code:
            |
            |                 *
            |            *
            |         *  |
            |       *    |y
            |    *       |
          --*------------+---
            |      x

    The area of a triangle is: A \,= \,\frac{1}{2}bh

    The area of this triangle is: . A \:=\:\frac{1}{2}xy . . . where y \:=\:x^2e^{-3x}

    So we have: . A \:=\:\frac{1}{2}x\cdot x^2e^{-3x}\quad\Rightarrow\quad A \:=\:\frac{1}{2}x^3e^{-3x}

    Differentiate: . A' \;= \;\frac{1}{2}x^2\!\cdot\!e^{-3x}(-3) + \frac{1}{2}\!\cdot\!3x^2\!\cdot\!e^{-3x} \;= \;\frac{3}{2}x^2e^{-3x}(-x + 1)

    Equate to zero: . \frac{3}{2}x^2e^{-3x}(-x + 1)\;=\;0


    Set each factor equal to zero and solve.

    . . \frac{3}{2}x^2\,=\,0\quad\Rightarrow\quad x\,=\,0 . . . This gives minimum area ... ha!

    . . e^{-3x} \,=\,0 . . . but \frac{1}{e^{3x}} can never equal zero.

    . . -x + 1 \:=\:0\quad\Rightarrow\quad x\,=\,1 . . . There!


    Maximum area occurs when \boxed{x \,= \,1} and y \,= \,1^2\!\cdot\!e^{-3} \,= \,\frac{1}{e^3}

    The maximum area is: . A \:=\:\frac{1}{2}(1)\left(\frac{1}{e^3}\right) \:=\:\boxed{\frac{1}{2e^3}\text{ square units}}

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  3. #3
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    thanks!

    Is that LaTeX that you used to format the text?

    If so, what's the tag for it in these forums? I know the other forums I use are , and they don't seem to work here
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  4. #4
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    Quote Originally Posted by Jacobpm64 View Post
    thanks!

    Is that LaTeX that you used to format the text?

    If so, what's the tag for it in these forums? I know the other forums I use are , and they don't seem to work here
    This is what you use.
    [tex] [/tex]
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