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Thread: Help with max and min.

  1. #1
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    Help with max and min.

    A right triangle has one leg on the x-axis.The vertex at the right end of that leg is at the point (3,0).The other vertex touches the graph of y=e^x.The entire triangle is to lie in the first quadrant.Find the max area of this triangle.

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  2. #2
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    Quote Originally Posted by jupiter92 View Post
    A right triangle has one leg on the x-axis.The vertex at the right end of that leg is at the point (3,0).The other vertex touches the graph of y=e^x.The entire triangle is to lie in the first quadrant.Find the max area of this triangle.

    Thanks.
    From what you have described, it would appear that your triangle has vertices:

    $\displaystyle (3, 0), (x, e^x), (x, 0)$ and I assume that $\displaystyle x < 3$.


    Therefore its base is $\displaystyle 3 - x$ and its height is $\displaystyle e^x$.


    So its area is given by the formula:

    $\displaystyle A = \frac{1}{2}(3 - x)e^x$.


    Since you want the maximum area:

    $\displaystyle \frac{dA}{dx} = \frac{1}{2}\left[(3 - x)e^x - e^x\right]$

    $\displaystyle = \frac{1}{2}e^x(3 - x - 1)$

    $\displaystyle = \frac{1}{2}e^x(2 - x)$.


    Setting the derivative equal to $\displaystyle 0$ and solving for $\displaystyle x$:

    $\displaystyle \frac{1}{2}e^x(2 - x) = 0$

    $\displaystyle e^x = 0$ or $\displaystyle 2 - x = 0$.


    But since $\displaystyle e^x > 0$ for all $\displaystyle x$, that means that only $\displaystyle 2 - x = 0$ is valid.

    Therefore $\displaystyle x = 2$.


    To check that this IS a maximum, find the second derivative and check that it is negative at $\displaystyle x = 2$.


    $\displaystyle \frac{d^2A}{dx^2} = \frac{1}{2}\left[e^x(2 - x) - e^x\right]$

    $\displaystyle = \frac{1}{2}e^x(2 - x - 1)$

    $\displaystyle = \frac{1}{2}e^x(1 - x)$


    At $\displaystyle x = 2$

    $\displaystyle \frac{d^2A}{dx^2} = \frac{1}{2}e^2(1 - 2)$

    $\displaystyle = -\frac{1}{2}e^2$.


    This function is negative at $\displaystyle x = 2$, so $\displaystyle x = 2$ does indeed maximise the area.


    So at $\displaystyle x = 2$:

    $\displaystyle A = \frac{1}{2}(3 - 2)e^2$

    $\displaystyle = \frac{1}{2}e^2$ square units.
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  3. #3
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    thanks so much.
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