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Math Help - The fireman's problem (optimization)

  1. #1
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    The fireman's problem (optimization)

    A fence a feet high is b feet from a high burning building. Find the length of the shortest ladder that will reach from the ground across the top of the fence to the building.

    I have the picture set up and now I'm stuck. I know that I have to express the ladder in terms of x and I'm pretty sure that it involves similar triangles. Thanks.
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  2. #2
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    Hello, elitespart!

    I found that Trig made this easier . . .


    A fence a feet high is b feet from a burning building.
    Find the length of the shortest ladder that will reach from the ground
    across the top of the fence to the building.
    Code:
                                  * R
                               *  |
                            *     |
                         * θ      |
                    Q * - - - - - * S
                   *  |     b     |
                *     |a          |
             * θ      |           |
          * - - - - - * - - - - - *
          P           U     b     T

    The fence is QU = a feet.
    It is UT = b feet from the building.
    . . Note that QS = b.

    The ladder is PR = L, which makes angle \theta with the ground.
    .Note that \angle RQS = \theta.

    In right triangle QUP\!:\;\;\csc\theta = \frac{PQ}{a}\quad\Rightarrow\quad PQ = a\csc\theta

    In right triangle RSQ\!:\;\;\sec\theta = \frac{QR}{b} \quad\Rightarrow\quad QR = b\sec\theta

    We have: . PR \;=\;L \;=\;a\csc\theta + b\sec\theta .[1]


    Minimize L, solve \frac{dL}{d\theta} = 0

    . . \frac{dL}{d\theta} \;=\;-a\csc\theta\cot\theta + b\sec\theta\tan\theta \;=\;0

    We have: . -a\cdot\frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\  sin\theta} + b\cdot\frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\  cos\theta} \;=\;0 . . \Rightarrow\quad \frac{b\sin\theta}{\cos^2\!\theta} - \frac{a\cos\theta}{\sin^2\!\theta} \;=\;0


    Multiply by \sin^2\!\theta\cos^2\!\theta\!:\;\;b\sin^3\!\theta - a\cos^3\!\theta \;=\;0 \quad\Rightarrow\quad b\sin^3\!\theta \;=\;a\cos^3\!\theta

    . . \frac{\sin^3\!\theta}{\cos^3\!\theta} \;=\;\frac{a}{b} \quad\Rightarrow\quad \left(\frac{\sin\theta}{\cos\theta}\right)^3 \;=\;\frac{a}{b} . . \Rightarrow\quad \tan^3\!\theta \;=\;\frac{a}{b} \quad\Rightarrow\quad \tan \theta \;=\;\sqrt[3]{\frac{a}{b}}


    We have: . \tan\theta \;=\;\frac{\sqrt[3]{a}}{\sqrt[3]{b}} \;=\;\frac{opp}{adj}

    Using Pythagorus, we have: . \begin{Bmatrix}opp &=& a^{\frac{1}{3}} \\ adj &=& b^{\frac{1}{3}} \\ hyp &=& \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}} } \end{Bmatrix}

    Then: . \csc\theta \;=\;\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} {a^{\frac{1}{3}}} \qquad \sec\theta \;=\;\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}


    Substitute into [1]: . L \;=\;a\cdot\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{a^{\frac{1}{3}}} + b\cdot\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} {b^{\frac{1}{3}}} . = \;a^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} + b^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}

    Factor: . L \;=\;\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)


    Therefore: . \boxed{L \;=\;\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}}

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  3. #3
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    Wow! Thank you so much. I really appreciate it.
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  4. #4
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    Re: The fireman's problem (optimization)

    i had the same question in my study guide :/ i couldn't get the answer :S can u explain more deeply?
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