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Maximum Fenced Area, One Side a Barn


Date: 10/16/2001 at 10:00:48
From: Molly Flamion
Subject: Algebra II

My question is about solving for the equation y = 70x - 2x squared.
What we had to do was find the maximum area of a fence if one side was 
a barn. Our class thought that it was going to be a square, but then 
we realized that it was going to be a rectangle, and the rectangle 
turned out to be two squares. Why did the rectangle turn out to be two 
equal squares?


Date: 10/16/2001 at 12:36:14
From: Doctor Greenie
Subject: Re: Algebra II

Hi, Molly -

This is a difficult question to answer; I can see why it turns out 
that way, but I'm not sure I can explain it to you.

Evidently you are finding the maximum rectangular area you can enclose 
with 70 units of fencing if one side of the area is a barn (and so 
needs no fencing). So you have a picture like this...

                     BARN
           -------------------------
               |               |
               |               |
             x |               | x
               |               |
               +---------------+
                     70-2x

The area is length times width:

    A = x(70-2x)

If we treat this as the equation of a parabola, we can find where the 
area is maximum by finding the vertex of the parabola; we can do this 
by completing the square. We start out as follows...

    y = x(70-2x)
    y = -2x^2+70x
    y = -2(x^2-35x+ ...) + ...
    y = -2(x-(35/2))^2 + ...

So we see the parabola opens downward and the maximum value is when 
x = 35/2. The fenced area is then 17.5 units wide and 35 units long.

Now let's compare this problem to the same problem without the barn:  
Find the dimensions of the rectangular field of maximum area that can 
be enclosed with 70 units of fencing. If we set up the problem the 
same way we did the first problem, we have a figure like this:

             35-x
        +-------------+
        |             |
        |             |
        |             |
      x |             | x
        |             |
        |             |
        +-------------+
             35-x

The area is again length times width:

    A = x(35-x)

If, as before, we treat this as the equation of a parabola and find 
the vertex by completing the square, we start out as follows...

    y = x(35-x)
    y = -x^2+35x
    y = -1(x^2-35x+ ...) + ...
    y = -1(x-(35/2))^2 + ...

So we see the parabola also opens downward, and the maximum value is 
again when x = 35/2. The fenced area is then 17.5 units long and 
17.5 units wide.

So, in both problems, the maximum area is obtained when the width is 
one-fourth of the total amount of fencing available. In the case 
without the barn, that means one-fourth of the fencing is used on each 
side of the area; in the case with the barn, it means that half of the 
fencing is available for the length of the rectangular area. So the 
rectangular area in the case with the barn is twice as long as it is 
wide - i.e., it is, as you put it, "two squares."

I hope this helps. Write back if you have any further questions on 
this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/16/2001 at 12:47:03
From: Doctor Peterson
Subject: Re: algebra II

Hi, Molly.

You correctly saw that symmetry is a key to this sort of problem; if 
you just had a rectangular fence alone, the square would give the 
maximum area because of its symmetry. Both "height" and "width" play 
the same role in the equations you write (in that case), so the 
solution should be one in which they have the same value. This is not 
always true, but is always a good guess.

          w
    +-----------+
    |           |        V = wh
    |           |
    |           |h       2w + 2h = P
    |           |
    |           |        V = (P/2 - h)h = P/2 h - h^2
    +-----------+

Now, in the problem you did, the variables do not play symmetrical 
roles; the length of the fence involves the height twice, but the 
width only once:

          w
    +-----------+       V = wh
    |           |h
    |           |       w + 2h = S
  ==+===========+==
                        V = (S - 2h)h = Sh - 2h^2
                         
But we can make the problem symmetrical, by reflecting the fence in 
the wall:

          w
    +-----------+
    |           |
    |           |2h
  ==+===========+==
    |           |
    |           |
    +-----------+

Maximizing the area enclosed by the fence and the wall will also 
maximize the doubled area enclosed by the fence and its reflection. 
But the latter is the same as the first problem I worked out; w and h 
are now fully symmetrical. Therefore, the maximum area is obtained 
when w and 2h are equal, which gives your 2:1 ratio.

This is an example of the fun we can have in math: transforming a new 
problem into an old one (in this case by reflecting) so that we can 
solve it without much extra thought. We try to avoid doing actual work 
as much as possible!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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