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Math Help - tricky optimization question

  1. #1
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    Angry tricky optimization question

    Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by aaasssaaa View Post
    Find the dimensions of the isosceles triangle of largest area that can be inscribed in a circle of radius r.
    First see the attached diagram.

    1. Use the Intersecting Chord Theorem to get:

    <br />
(b/2)^2=x(2r-x)<br />

    This is a quadratic in x, which we may solve using the quadratic formula:

    <br />
x=r \pm \sqrt{4r^2-b^2)}<br />

    Now we need the -ve sign in this so:

    <br />
x=r(1 - \sqrt{1-(b/(2r)^2)})<br />

    2. The area of the triangle is:

    <br />
A = \frac{b\ h}{2}=\frac{b\ (2r-x)}{2}=\frac{b(2r-r \left[ 1-\sqrt{1-(b/2r)^2}\right] )}{2}<br />

    3. Now to find the isosceles triangle which maximises this area you need for
    solve:

    \frac{dA}{db}=0

    and if there are multiple solutions sort through them to identify which corresponds
    to a maximum, and then find the other sides of the triangle.

    RonL
    Attached Thumbnails Attached Thumbnails tricky optimization question-gash.jpg  
    Last edited by CaptainBlack; November 11th 2006 at 01:46 AM.
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  3. #3
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    Hello, aaasssaaa!

    Here's my approach to this problem . . .


    Find the dimensions of the isosceles triangle of largest area
    that can be inscribed in a circle of radius r.
    Code:
                    A
                  * * *
              *     :     *
            *       :       *
           *       r:        *
                    :
          *         :         *
          *         *         *
          *         : \       *
                   y:   \r
           *        :     \  *
           B* - - - + - - - *C
              *        x  *
                  * * *

    Draw line segments AB and AC.

    We will maximize the area of \Delta ABC
    . . which has base 2x and height y + r.

    The area of the triangle is: . A \:=\:\frac{1}{2}(2x)(y + r) \:=\:x(y + r) [1]

    Using Pythagorus, we have: . x^2 + y^2\,=\,r^2\quad\Rightarrow\quad y = \sqrt{r^2 - x^2} [2]

    Substitute [2] into [1]: . A \;= \;x\left[(r^2 - x^2)^{\frac{1}{2}} + r\right]


    Differentiate: . A' \;=\;x\left[\frac{1}{2}(r^2 - x^2)^{-\frac{1}{2}}(-2x)\right] + (r^2-x^2)^{\frac{1}{2}} + r

    Equate to zero: . \frac{-x^2}{\sqrt{r^2-x^2}} + \sqrt{r^2-x^2} + r \;= \;0

    Multiply by \sqrt{r^2-x^2}\!:\;\;-x^2 + r^2 - x^2 + r\sqrt{r^2-x^2} \:=\:0

    Then we have: . r\sqrt{r^2-x^2} \:=\:2x^2 - r^2

    Square both sides: . r^2(r^2-x^2) \:=\:(2x^2-r^2)^2\quad\Rightarrow\quad r^4 - r^2x^2\:=\:4x^4 - 4r^2x^2 + r^4

    . . 4x^4 - 3r^2x^2\:=\:0\quad\Rightarrow\quad x^2(4x^2 - 3)\:=\:0


    If x^2=0, we have x = 0 which produces minimum area . . . ha!

    If 4x^2-3 = 0, we have: . \boxed{x = \frac{\sqrt{3}}{2}r}

    Substitute into [2] and we get: . \boxed{y = \frac{r}{2}}


    Therefore, the triangle has: . \begin{array}{cc}\text{base} \\ \text{height}\end{array}\begin{array}{cc}=\\=\end{  array} \begin{array}{cc}2x\\y+r\end{array} \begin{array}{cc}=\\=\end{array}\begin{array}{cc}\  sqrt{3}r \\ \frac{3}{2}r\end{array}


    Note: The triangle happens to be equilateral,
    . . . . .which you may have suspected all along.

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  4. #4
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    Why did you need to find the negative sign of the quadratic?
    Abd how did it turn out to be

    Now we need the -ve sign in this so:

    x=r(1 - \sqrt{1-(b/(2r)^2)})
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by liquidanubis3 View Post
    Why did you need to find the negative sign of the quadratic?
    Abd how did it turn out to be

    Now we need the -ve sign in this so:

    x=r(1 - \sqrt{1-(b/(2r)^2)})
    Please quote what you are referring to.

    (if it is in solving the quadratic resulting from the intersecting chord theorem its because we want x to be the smaller or the two roots)

    CB
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