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Math Help - Application of maxima and minima

  1. #1
    Junior Member Kaloda's Avatar
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    Application of maxima and minima

    Find the length of the longest rod which can be carried horizontally around a corner from a corridor 8m wide into one 4m wide. (Without involving angles if possible)
    VIEW the attachment for the figure.
    Attached Thumbnails Attached Thumbnails Application of maxima and minima-max.png  
    Last edited by Kaloda; September 30th 2012 at 08:43 AM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Application of maxima and minima

    Let:

    L^2=(8+x)^2+(4+y)^2

    By similarity, we have:

    \frac{y}{8}=\frac{4}{x}

    Can you proceed from here?
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  3. #3
    Junior Member Kaloda's Avatar
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    Re: Application of maxima and minima

    I already did but it gave me a wrong answer. Maybe something is wrong with my solution. I'll check it.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Application of maxima and minima

    What did you get for your critical value?
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  5. #5
    Junior Member Kaloda's Avatar
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    Re: Application of maxima and minima

    4*2^(2/3) on my second soltn and 8 on the first.

    Btw, w/c function must I use? The one above or this one (L=a+b):
    L(x)=a+b=sqrt(x^2+64)+sqrt(y^2+16) ??
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  6. #6
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    Re: Application of maxima and minima

    Hello, Kaloda!

    This is a classic problem . . .


    Find the length of the longest rod which can be carried horizontally around a corner
    from a corridor 8m wide into one 4m wide.

    Code:
          |       |
        A *       |
          | *     |
         y|   *   |
          |     * | P
        D + - - - * - - - - - - - -
          |   4   : *
          |       :   *
         8|      8:     *
          |       :       *
          |       :         *
          * - - - + - - - - - * - - - -
        C     4   E     x     B
    Let L \,=\,AB be the length of the rod.

    . . Then: . L \;=\;\sqrt{(x+4)^2 + (y+8)^2} .[1]


    We see that \Delta PEB \sim \Delta ADP.

    . . Hence: . \frac{x}{8} \,=\,\frac{4}{y} \quad\Rightarrow\quad y \:=\:\frac{32}{x}


    Substitute into [1], set L' \,=\,0, and solve for maximum L.
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  7. #7
    Junior Member Kaloda's Avatar
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    Re: Application of maxima and minima

    Okay. That's exactly how I solve the problem but I obtain a wrong answer! Maybe I'm missing something.
    And ow, the book hints that this length (max) is the minimum of certain lengths.
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  8. #8
    MHF Contributor MarkFL's Avatar
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    Re: Application of maxima and minima

    I get (substituting for y into the square of the length) the following as the critical value:

    x=4\sqrt[3]{2}

    I began with:

    L^2=(8+x)^2+(4+y)^2

    L^2=(8+x)^2+\left(4+\frac{32}{x} \right)^2

    L^2=(8+x)^2+16\left(1+\frac{8}{x} \right)^2

    Implicitly differentiating with respect to x, we have:

    2L\cdot\frac{dL}{dx}=2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)

    Since we must have 0<L, to find the extrema, we may equate:

    2(8+x)+32\left(1+\frac{8}{x} \right)\left(-\frac{8}{x^2} \right)=0

    (8+x)-128\left(1+\frac{8}{x} \right)\left(\frac{1}{x^2} \right)=0

    (8+x)-128\left(\frac{1}{x^2}+\frac{8}{x^3} \right)=0

    Multiplying through by x^3 and distributing, we find:

    x^4+8x^3-128x-1024=0

    Factor:

    (x+8)(x^3-128)=0

    Discarding the negative root, we have:

    x^3=128=2\cdot2^6

    x=2^2\cdot\sqrt[3]{2}=4\sqrt[3]{2}

    I suspect you have set it up like Soroban did, so your critical value may be different from what I have.
    Last edited by MarkFL; September 30th 2012 at 10:44 AM.
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  9. #9
    Junior Member Kaloda's Avatar
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    Re: Application of maxima and minima

    Oh yeahhh.! TY for the replies. I got the correct answer but I didn't realized that it just has a different form from the one given by the book.
    Didn't know it til I checked their numerical values. I'm such a f*ol!
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