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Thread: Need help understanding these types of problems: fencing 2 side-by-side areas

  1. #1
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    Need help understanding these types of problems: fencing 2 side-by-side areas





    Farmer Ted wants to build two pastures side by side with shared wall between them. Ted has $4800 to spend. The fencing around the perimeter costs $3 per foot and the
    shared fencing costs $2 per foot. What is the max area Farmer Ted can enclose?
    What are the dimensions of the pastures?

    I have some work already done that attempts to answer the problem

    3x+2y=4800
    -3(x+y=5)
    -----------
    2y=4800
    -3y=-15
    ----------
    -y= 4785

    Im pretty sure this is solvable using a quadratic equation because the notes i took in class are over a similar problem but its set up differently so the steps to solving it are different as well. Im not exactly sure what to do with the -y=4785 that i got or if its even set up right.

  2. #2
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    Quote Originally Posted by Matts View Post




    Farmer Ted wants to build two pastures side by side with shared wall between them. Ted has $4800 to spend. The fencing around the perimeter costs $3 per foot and the
    shared fencing costs $2 per foot. What is the max area Farmer Ted can enclose?
    What are the dimensions of the pastures?

    I have some work already done that attempts to answer the problem

    3x+2y=4800
    -3(x+y=5)
    -----------
    2y=4800
    -3y=-15
    ----------
    -y= 4785

    Im pretty sure this is solvable using a quadratic equation because the notes i took in class are over a similar problem but its set up differently so the steps to solving it are different as well. Im not exactly sure what to do with the -y=4785 that i got or if its even set up right.
    Let the total field be L (ft) * W (ft).

    Let the length of the shared fence be W (ft).

    Then

    3 * 2 * (L + W) + 2 * W = 4800 → 3L + 4W = 2400 → W = 600 - 0.75*L .......................(1)

    A = L * W ............. Using (1), we get

    A = L * (600 - 0.75 * L) ...... This is a quadratic and continue.....
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Quote Originally Posted by Denis View Post
    Or since a square produces maximum area,
    you can use side = L, thus perimeter 4L.

    You can then forget about the shared fence:
    instead, use 3 sides @ $3 and other side @ $5.
    You can't forget about the [cost of the] shared fence. The solutions don't involve pastures that are square-shaped.

    That is, the pastures turn out not to be square-shaped.

  4. #4
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    Quote Originally Posted by Denis View Post
    Or since a square produces maximum area,
    you can use side = L, thus perimeter 4L.

    You can then forget about the shared fence:
    instead, use 3 sides @ $3 and other side @ $5.
    That logic will not work here because of the fixed (but variable) cost involved.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
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    Quote Originally Posted by Denis View Post
    Ahhh shucks......>|_
    And to think you were outstanding in your field.

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    Quote Originally Posted by Denis View Post
    Or since a square produces maximum area,
    you can use side = L, thus perimeter 4L.

    You can then forget about the shared fence:
    instead, use 3 sides @ $3 and other side @ $5.
    Corner time mon ami.

    I have no idea why a problem of optimization under constraint is posed in algebra, but constraints may, and usually do, affect the answer. Moreover the problem is badly worded because, as given, there is no indication what, if any, constraints are placed on the shape of either pasture. I don't know how to solve this problem without limitations on the geometry; it may require calculus of variations or some even more advanced technique.

    If we set up two equal semicircles joined by a wall along the diameter, the cost equation is

    [tex]3 * \pi d + 2d = 4800 \implies d = \dfrac{4800}{3 \pi + 2} \implies[/tex]

    the combined area is

    [tex]\pi \left ( \dfrac{2400}{3 \pi + 2} \right )^2 \approx 138,636\ sq.\ ft.[/tex],

    which considerably exceeds the area formed by rectangles. So I am quite confident the problem, at least as stated, is beyond the power of algebra to solve. I have an intuition, which I cannot prove, that the correct answer involves symmetric semi-ellipses divided along the major axis.
    Last edited by JeffM; 02-19-2018 at 06:41 PM.

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    Quote Originally Posted by JeffM View Post
    Corner time mon ami.

    I have no idea why a problem of optimization under constraint is posed in algebra, but constraints may, and usually do, affect the answer. Moreover the problem is badly worded because, as given, there is no indication what, if any, constraints are placed on the shape of either pasture. I don't know how to solve this problem without limitations on the geometry; it may require calculus of variations or some even more advanced technique.

    If we set up two adjacent semicircles of equal diameter divided by a diameter fence, the cost equation is

    [tex]3 * \pi d + 2d = 4800 \implies d = \dfrac{4800}{3 \pi + 2} \implies[/tex]

    the combined area is

    [tex]\pi \left ( \dfrac{2400}{3 \pi + 2} \right )^2 \approx 138,636[/tex],

    which considerably exceeds the area formed by rectangles.
    I concur. I saw the OP's attempt at the solution (3x + 2y = 1800) and inferred (never assumed) that a rectangle was "suggested" in class.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  8. #8
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    Also, there is this:


    Quote Originally Posted by Matts View Post

    Farmer Ted wants to build two pastures side by side with shared wall [tex] \ \ \ \ [/tex] Here it is called a "wall." One would think there would be no fencing.

    between them. Ted has $4800 to spend. The fencing around the perimeter costs $3 per foot and the

    shared fencing [tex] \ \ \ \ [/tex] But here it is called "shared fencing."

    costs $2 per foot.

    .
    Last edited by lookagain; 02-19-2018 at 05:55 PM.

  9. #9
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    Quote Originally Posted by Denis View Post
    Well, certainly looks like a rectangle was intended (as Subhotosh says);
    so 400 by 300 = 120,000 sq.ft.

    800*$3 + 600*$3 + 300*$2 = $4800

    Corner??
    138 > 120

  10. #10
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    Quote Originally Posted by Denis View Post
    You've lost me Jeff: what meanest thou?
    Farmer Ted is not too smart. He can get 238K sq ft with just a bit of cleverness. No one told him that sheep care whether their pastures are rectilinear. You are misleading farmer Ted, and he needs all the help he can get.

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