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Perimeter vs. Area

  1. Aug 18, 2008 #1

    A total of x feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of x?


    My first guess is x^2 / 9 if you assume the yard is square. I'm a little unsure of my answer since for either the horizontal or vertical (depending on how you set up your yard) you only have to fence one side of the yard, thus it may be beneficial to extend that side since you are required to fence less of it and increase overall area. Can anyone please explain their answer in comparison to mine?

    What I did:
    A= Area
    X= Fencing in feet
    L= Length
    W= Width

    A= L*W
    X = 2W + L

    X = 2W + A/W
    XW - 2W^2 = A

    Which takes a maximum at W = X/4

    Thus area should be maximized at L = X/2, W= X/4 thus maximum area is X^2/8

    This is for the subject Math GRE (Question 13 on the prep exam), how would any of you do this quickly? My way seemed to take more time then I would want to spend on something like this.
  2. jcsd
  3. Aug 18, 2008 #2


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    moo5003, your approach looks good, although I did not examine the latter steps in any detail. Apparant is that you have a quadratic term with a negative sign, which is good because you will want to obtain a parabola with the vertex at a maximum point.
  4. Aug 18, 2008 #3


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    Your answer is correct. And you have done it quickly: that is the way pretty much anyone would show the solution. The only step you didn't show was

    d/dW (XW - 2W^2) = d/dW (A)

    --> X - 4W = 0

    and you have the rest.

    The only situation for a "fencing problem" where the maximal area/minimal perimeter solution is a square is when the fencing material is used only for the full perimeter of the enclosure or where any partitioning of the enclosure is performed with four-fold symmetry (equal numbers of north-south and east-west partitions). Other arrangements disrupt the simplest symmetry of the enclosure and produce "extremal" solutions which are not squares.
    Last edited: Aug 18, 2008
  5. Jan 24, 2011 #4
    This is another method I came up with that I believe is faster.

    X = 2W + L; where X is a constant.

    which can also be rewritten as:

    L = -2W + X; Think of this as y=mx+b (so L-axis is y-axis and W-axis is x-axis)

    X is L-intercept) and 2W is the W-intercept)

    Then the max of a rectangle with one vertex on this line is at (W, X/2) (middle of the line)

    Which means L = X/2 and then use X = 2W + L to find W (which comes out to W = X/4)

    Finally: A = L*W = (X/2)(X/4) = (x^2)/8

    Sorry if my notation got confusing in the middle.
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