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Math Help - Maximum Area

  1. #1
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    Post Maximum Area

    Thanks!
    Last edited by Blasnmt23; May 4th 2010 at 04:28 PM.
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  2. #2
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    Hello, Blasnmt23!

    I used a simpler approach . . .


    The line y \:=\: mx + b intersects the parabola  y = x^2 in points A and B.
    Find the point C that maximizes the area of the triangle.
    Code:
                      |
       o              |          ....♥ B
                      |  ....*::::::*
                  ....*::::::::::::*
           ....*::::::|:::::::::::*
      A ♥:::::::::::::|::::::::::*  o
            *:::::::::|:::::::::*
         o      *:::::|::::::::*   o
          o         *:|:::::::*   o
            o         | *::::*  o
                o     |     ♥
      - - - - - - - - o - - C - - - - - - -
                      |

    Point A has coordinates (a,a^2)
    Point B has coordinates (b,b^2)
    Point C has coordinates (x,x^2)


    The area of \Delta ABC is given by: . A \;=\;\frac{1}{2}\left|\begin{array}{ccc}1&1&1 \\ a&b&x \\ a^2&b^2&x^2\end{array}\right|


    We have: . A \;=\;\tfrac{1}{2}\bigg[(bx^2 - b^2x) - (ax^2 -a^2x) + (ab^2-a^2b)\bigg]

    . . . . . . . . A \;=\;\tfrac{1}{2}\bigg[(b-a)x^2 - (b^2-a^2)x + ab(b-a)\bigg]


    Maximize: . \frac{dA}{dx} \;=\;\tfrac{1}{2}\bigg[2(b-a)x - (b^2-a^2)\bigg] \;=\;0

    . . . . . . . 2(b-a)x \:=\:b^2-a^2 \quad\Rightarrow\quad x \;=\;\frac{(b-a)(b+a)}{2(b-a)} \quad\Rightarrow\quad x \;=\;\frac{a+b}{2}


    The coordinates of point C are: . \left(\frac{a+b}{2},\;\left[\frac{a+b}{2}\right]^2\right)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Fascinating!

    Point C is exactly "halfway between" points A and B.

    Moreover, C is the point at which the tangent is parallel to the secant AB.

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