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Need ur guys help again

  1. Apr 21, 2004 #1
    Need ur guys help again ....

    Heres the question .........

    A Cattle Farmer wants to build a rectangular fenced enclosure divided into 3
    rectangular pens. A total lenght of 120m of fencing material is available. Find the overall dimensions of teh enclosure that will make the total area a maximum. Explain and justify your reasoning.

    Any help is needed and will be gretaly appreciated .. Thx for ur time ..
     
  2. jcsd
  3. Apr 22, 2004 #2

    uart

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    It's an incompletely specified problem. If you had a requirement that each of the three rectangular sub-enclosures were of equal area (or even of some minumum size) then you'd be ok.

    As it stands there is a solution (an actual maximum total area) but it is not a practical one, as the area of two of the sub-enclosuers vanishes to zero. This is not an error in the maths or the approach, but an error in the problem description. My solution will yield a total area approaching 600m^2. I'll dont you'll find a solution that can exceed this.
     
  4. Apr 22, 2004 #3

    honestrosewater

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    When the length of both inner fences = 0m, i.e., all of the 120m of fence is in the perimeter, the area = ab, where a + b = 60.
    Wouldn't a=30 give the maximum area (900m^2), making 900m^2 the limit?
    Perhaps I missed something or didn't take into account that the area is a limit of a limit.

    Also, unless a rectangle is defined here to have unequal sides, can't the outer pen or inner pens be square, i.e., a rectangle with equal sides. I mean, shouldn't the question say "unequal sides" or "a[not=]b" etc. somewhere? And since it doesn't, you can reach your limit of 900m^2?

    Happy thoughts,
    Rachel
     
  5. Apr 22, 2004 #4

    honestrosewater

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    Okay, I think I see, at least conceptually. It does make a difference if your inner fences are parallel or perpendicular, yes?
    If your inner fences are perpendicular to each other, as the areas of two inner pens approach zero, the length of one inner fence can approach zero, while the length of the other inner fence equals the length of the shortest outer fence.
    If they are parallel, both must have the same (longer) length.
    ??
    Rachel
     
  6. Apr 22, 2004 #5

    uart

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    Ok, I'm assuming that all three sub-enclosures must be rectangular not just in the limiting case but also for the finite cases from which they are derived. I think that's fair enough as this is a supposedly a "practical" problem. That means that you can't make the inner boundaries of the pens arbitrary, they must divide the outer pen into three genuine rectangles.

    For that reason I choose a "T" shaped inner boundary to divide the outer rect into three sub-rectangles. You can then shrink the vertical part of the "T" to an arbitrarily small length while maintaining all three inner pens as rectangles at all times. The resultant perimeter shrinks towards a limit of 3a + 2b, where "a" and "b" are the side lengths of the outer rectangle.

    So the problem becomes to maximize a*b while maintaining 3a+2b=120. I use the method of Lagrange multipliers to get a=20 and b=30 and a limiting total area of 600 m^2.

    I realize that if you allow the inner pens to be non-rectangular then you can get all 120m of perimeter into the outer pen in the limit but I did not count this as valid. An example of this would be to make the outer pen a square and to make the inner boundaries say just little "L" shaped fences on two of the corners. As you shrink the two corner pens toward zero then you get the full 120m in the outer perimeter and a resultant area of 900m^2. The problem with this is that the two corner pens are rectangular as required but the third pen is non rectangular. See what I mean?

    So I stand by my solution. Make the inner boundary a "T" and you can make the total area arbitrarily close to 600m^2 while keeping all three inner pens rectangular at all times.

    BTW. Yes a square is a perfectly valid rectangle. This is just as true as saying that a cow is an animal. Someone might argue that a cow is actually a mammal (more specific) and that is true, but in no way does that prevent a cow from also being an animal right. The same is true for classification of shapes.
     
  7. Apr 22, 2004 #6

    honestrosewater

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    Hello, Uart :)
    Your "T" = my "perpendicular inner fences". I don't know what "the method of Lagrange multipliers" means, but no matter, I will trust that you know what it means ;) So if the outer pen is square, 3a + 2b = 120 becomes 5a = 120 ...(120/5)^2 = 576 and I don't regret trusting you. And if the inner fences have to be parallel, maximize a*b while maintaining 4a + 2b = 120. And does this become a = 15, b = 30? and, generally, xa = yb = t/2, where t = total perimeter.

    I agree about the "divided into three rectangular pens"- I wasn't suggesting anything like the "L-shaped" corners, I just wasn't thinking about the problem properly- that both inner fences can't equal zero. I have a small problem with this- for "T", when the vertical line vanishes, and so is not counted as a line in 3a+2b=120, the horizontal line is *on* the perimeter. My memory is saying "but the ratio still holds", but for some reason, I don't like it. Guess I have to grow ;)
    Happy thoughts
    Rachel
     
  8. Apr 22, 2004 #7

    uart

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    Rachel, consider the following arrangment.

    Outer enclusure dimensions : 30m by 19.9m

    Inner "T" boundary : 19.9m by 0.3m. With the 19.9m section running right across the enclosure at a distance of 0.3m from one of the 19.9m outer boundaries. And the 0.3m part of the "T" ofcourse used to divide this 19.9 by 0.3 sub-rectangle into two parts (I'll choose two equal parts though that is arbitrary).

    So we have three inner rectangles exactly as required. They are
    a) 29.7 by 19.9
    b) 9.95 by 0.3
    c) 9.95 by 0.3

    The total perimiter is 2*30 + 3*19.9 + 1*0.3 = 120m, exactly as required.

    The outer enclosure is also a rectangle exactly as required. The total area is 30*19.9 = 597m^2

    What's wrong with that solution?

    And you can go further, you can get as close as you like to 600m^2. (Though it's true you can never get exactly 600m^2, as at that limit two of the inner rectangles disappear).
     
  9. Apr 23, 2004 #8

    honestrosewater

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    Okay, I see.
    Thanks
    Rachel
     
  10. Apr 27, 2004 #9
    If i showed u guys a diagram woudl that work ??? I need to get teh solution to this by tommrow so any of ur guys help woudl be greatly aprreeciated .. If u think u can help me leve ur email address and i will send u a scan of the question with teh diagram ... i woudl put on here but its says teh dimenisons are to large so ... LMK Thx
     
  11. Apr 28, 2004 #10

    uart

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    The solution to the problem has been done and is posted above. If you have more details of the problem description that would require a different solution then just post them here.

    There is nothing to revisit unless the problem definition is tightned. Just provide details such as :

    Do the internal boundaries have to be parallel ?
    Do the sub-enclosures have to have equal area ?
    Do the individual sub-enclosures need to have a minimim area?
     
    Last edited: Apr 28, 2004
  12. Apr 28, 2004 #11
    The 3 rectangles have 3 equal areas .. mayeb this helps ?
     
  13. Apr 28, 2004 #12

    HallsofIvy

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    Yes, that helps- although it is still not completely determined. I'm going to assume that there is a large rectangle, with sides of length a and b, and the three pens are formed by putting two internal fences of length a (and, of course, parallel to the sides of length a). We will then need two fences of length b and four fences of length a: the total fencing needed is 2b+ 4a= 120 m. The total area (neglecting any area taken up by the fences) is ab. From 2b+ 4a= 120, b= 60- 2a so we want to maximize a(60- 2a)= 60a- 2a<sup>2</sup>= -2(a<sup>2</sup>- 30a).

    We can "complete the square" by adding and subtracting (-30/2)<sup>2</sup>= 225: the area is -2(a<sup>2</sup>- 30a+ 225- 225)= -2(a- 15)<sup>2</sup>+ 450.
    Since a square is never, negative, is a is anything other than 15, that will be 450 MINUS something. It will have its largest value, 450 square meters, when a= 15 meters and b= 60-2(15)= 30 meters.

    That's probably the solution intended.

    HOWEVER!! Rectangles just aren't efficient! It's well known that a circle includes the largest area for a given perimeter (and can be proved using "calculus of variations"). If we construct a circle of radius r and then run 3 fences from the center of the circle to the circumference, we will need 2pi r + 3r length of fence and enclose pi r<sup>2</sup> area.

    From 2pi r+ 3r= (2pi+ 3)r= 120, we get r= 120/(2pi+ 3)= 13 meters, approximately, and then the area is 2(pi)(13)<sup>2</sup>= 525 square meters.

    That's the REAL solution (ignoring that namby-pamby rectangle stuff)!
     
  14. Apr 29, 2004 #13

    uart

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    It's interesting that a circular enclosure is not actually the most efficient shape once you start requiring boundaries. When you start dividing it into regions you get an interesting dilemma. While straight boundaries like in a rectangle are inefficient (in terms of perimeter for a given area) they are however necessary to efficiently adjoin boundaries so that more than one sub-enclosure can share a common boundary. So you get a solution at has straight edges on the inner boundary and curved edges on the outer boundaries. Once you're forced to go with some straight edges then the overall most efficient shape does deviate from plain circle. I dont know the global "maximum area shape" out of all possible candidates but I suspect it would be somewhat shaped like a clover leaf.

    An example (not necessarily the best) can be constructed as follows:

    1. Start with an equilateral triangle with vertices a, b and c. (The triangle is just part of the construction and doesn't any fences yet, ok.)

    2. Mark the mid points of edges ab, bc and ca as well as the center of the triangle (that is, the point of intersection of lines connecting vertices to their opposite mid points).

    3. Build your first three fence segments from each of the vertices (a,b,c) into the center of the triangle. These are the shared inner boundaries between the enclosures.

    4. From the 3 midpoints of the edges of the triangle construct 3 semi-circular fence segments ending at the vertices of the respective edge. These are the non-shared outer boundaries.

    The resulting enclosure is a kind of a clover leaf shape with three equal area sub-enclosures. You can work out the dimension and hence area for a total fence length of 120m and turns out you can get about 559m^2 with this design.
     
    Last edited: Apr 29, 2004
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