Results 1 to 3 of 3

Math Help - Optimization Problem

  1. #1
    Junior Member NAPA55's Avatar
    Joined
    Mar 2008
    Posts
    58

    Optimization Problem

    A builder must completely fence in 3 adjacent rectangular lots along a roadway. Each lot must have an area of 675 square metres. Fencing along the roadway costs $25/metre while side and back fencing costs $10/metre. Find the dimensions of each lot that minimize the total cost of fencing.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    645
    Hello, NAPA55!

    The answers come out rather ugly . . .


    A builder must completely fence in 3 adjacent rectangular lots along a roadway.
    Each lot must have an area of 675 square metres.
    Fencing along the roadway costs $25/metre while side and back fencing costs $10/metre.
    Find the dimensions of each lot that minimize the total cost of fencing.
    Code:
               x         x         x
          * = = = = * = = = = * = = = = *
          |         |         |         |
          |         |         |         |
         y|        y|         |y        |y
          |         |         |         |
          |         |         |         |
          * - - - - * - - - - * - - - - *
               x         x         x

    There are 3x m of fencing along the road.
    At $25/m, its cost is: . (25)(3x) \:= \:75x dollars.

    There are 3x m of back fencing and 4y m of side fencing.
    At $10/m, its cost is: . 10(3x + 4y) \:=\:30x + 40y dollars.

    The total cost is: . C \;=\;75x + (30x + 40y) \;=\;105x + 40y dollars. .[1]


    Each lot has an area of 675 mē: . xy \:=\:675\quad\Rightarrow\quad y \:=\:\frac{675}{x}

    Substitute into [1]: . C \;=\;105x + 40\left(\frac{675}{x}\right) \quad\Rightarrow\quad C \;=\;105x + 25,000x^{-1}

    And that is the function we must minimize . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member polymerase's Avatar
    Joined
    May 2007
    From
    Sydney
    Posts
    267
    Just to finish off what Soroban started...

    Note: Remember your domain C =105x + 25,000x^{-1}\;\;\;\;0< x <\infty

    \frac {dC}{dx}=105-25000x^{-2}, set the derivative equal to zero...

    25000x^{-2}=105\to x^2=\frac{5000}{21}\to x=50\sqrt{\frac{2}{21}}. (x cannot be negative, as denoted by our domain, therefore we take the positive).

    Sub x back in and you should get y=\frac{27}{2}\sqrt{\frac{21}{2}}

    Therefore the dimension of each lot must be 50\sqrt{\frac{2}{21}} x \frac{27}{2}\sqrt{\frac{21}{2}}to minimize the cost of fencing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] optimization problem
    Posted in the Calculus Forum
    Replies: 12
    Last Post: October 6th 2011, 03:47 PM
  2. Optimization problem
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: March 4th 2011, 10:58 PM
  3. Optimization Problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2009, 07:41 PM
  4. help!!!- optimization problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 13th 2008, 04:59 PM
  5. Optimization Problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: April 8th 2008, 05:55 PM

Search Tags


/mathhelpforum @mathhelpforum