Results 1 to 9 of 9
Like Tree3Thanks
  • 1 Post By MarkFL
  • 1 Post By Soroban
  • 1 Post By MarkFL

Thread: Application of maxima and minima

  1. #1
    Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    77

    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; Sep 30th 2012 at 08:43 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Application of maxima and minima

    Let:

    $\displaystyle L^2=(8+x)^2+(4+y)^2$

    By similarity, we have:

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

    Can you proceed from here?
    Thanks from Kaloda
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    77

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Application of maxima and minima

    What did you get for your critical value?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    77

    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) ??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848

    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 $\displaystyle L \,=\,AB$ be the length of the rod.

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


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

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


    Substitute into [1], set $\displaystyle L' \,=\,0$, and solve for maximum $\displaystyle L.$
    Thanks from Kaloda
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    77

    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    2,005
    Thanks
    747

    Re: Application of maxima and minima

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

    $\displaystyle x=4\sqrt[3]{2}$

    I began with:

    $\displaystyle L^2=(8+x)^2+(4+y)^2$

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

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

    Implicitly differentiating with respect to $\displaystyle x$, we have:

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

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

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

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

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

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

    $\displaystyle x^4+8x^3-128x-1024=0$

    Factor:

    $\displaystyle (x+8)(x^3-128)=0$

    Discarding the negative root, we have:

    $\displaystyle x^3=128=2\cdot2^6$

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

    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; Sep 30th 2012 at 10:44 AM.
    Thanks from Kaloda
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    77

    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Oct 12th 2011, 06:01 AM
  2. Application of maxima and minima
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 7th 2011, 05:43 PM
  3. Application of Maxima and Minima
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 9th 2010, 06:54 AM
  4. Application: Maxima and Minima
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 7th 2008, 10:29 AM
  5. Application of maxima and minima
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Sep 23rd 2008, 07:46 AM

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum