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Math Help - 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:

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


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


    So its area is given by the formula:

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


    Since you want the maximum area:

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

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

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


    Setting the derivative equal to 0 and solving for x:

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

    e^x = 0 or 2 - x = 0.


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

    Therefore x = 2.


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


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

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

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


    At x = 2

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

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


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


    So at x = 2:

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

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