Enclosing land by fence pieces

Question

You have one 44-meter piece of fence and 48 one-meter pieces of fence. Those fences are straight and cannot be bent. What is the biggest area you can enclose with those fences on a two dimensional plane?

Answer 1

It is obvious that the solution shape is a convex polygon, probably a strictly convex one. Fix the 44-meter piece and consider the "loop" formed by the shorter pieces.

Take any three adjacent short pieces. Keep their endpoints fixed and connect them with a line $L$. This quadrilateral must have maximal area among those with the line $L$ as one edge and three other edges of length one. In this "variational subproblem" two points (those between one-meter pieces) are free to move. The solution to this subproblem is the symmetric one where the middle short piece is parallel to the long one. (I do not have a rigorous proof right now, but this should be easy enough to check.) This in particular implies that the two angles at the ends of the middle short piece are equal.

What we learn from the above is that any two adjacent angles between one-meter pieces must be equal. Thus they are all equal. Consequently all the 49 points lie one a circle.

It should be a straightforward calculation to find the area of this polygon.

Answer 2

As mentioned by other posters, in order to maximize the area, the points where the fences meet should lie on a circle. See the following reference. So you get something like the following picture, with, of course, $48$ segments of length $1$:

Note that the area of the enclosed region must then be:

$$A = 48 \left(\frac{1}{2} r^2 \sin \alpha\right) - \frac{1}{2} r^2 \sin (48\alpha)$$

And you can get the following system of equations:

$$ 2r \sin\left(\frac{\alpha}{2}\right) = 1$$

$$ 2r \sin\left(\frac{48\alpha}{2}\right) = 44$$

Solving for $r$ in both:

$$r = \frac{44}{2\sin(24\alpha)} = \frac{1}{2\sin\left(\frac{\alpha}{2}\right)}$$

$$44\sin\left(\frac{\alpha}{2}\right) = \sin(24\alpha)$$

Using Wolfram Alpha we get:

$$\alpha \approx 0.0298496$$

$$r \approx 33.502530$$

$$A \approx 248.1$$

Edit: As a follow-up after reading Dakvs's answer, if you allow only using a portion of the 44 meter segment, you can increase the area. So instead our picture looks like:

And we want to maximize:

$$A = 48 \left(\frac{1}{2} r^2 \sin \alpha\right) - \frac{1}{2} r^2 \sin (48\alpha)$$

Subject to the constraints:

$$ 2r \sin\left(\frac{\alpha}{2}\right) = 1$$

$$ 2r \sin\left(\frac{48\alpha}{2}\right) = B$$

$$ 0 \leq B \leq 44$$

I used Sage to find the area for all integer values of $B$, which is plotted below:

Since the maximum appears to occur for $B \in (29,32)$, I looked for the value of $B$ in this range to the nearest thousandth that maximizes $A$, and found the maximum occurs at approximately:

$$B \approx 30.562$$

$$A \approx 366.562$$

Answer 3

I'm not sure if it's maximal but this is my solution: you can achieve 288m^2 by creating a 12x24 square using only 24 meters of the 44m piece as one of the sides

Answer 4

I would guess that you should place the 48 $1m$ pieces in a circular arc over the $44m$ piece, since the continuous equivalent is precisely a section of a circle. At this point I am unable to give an exact formula for the resulting area, but I am sure it will be maximal.

Answer 5

A slightly different approach: If we make the substitution $\alpha = 2\arcsin\left(\frac{1}{2r}\right)$ in the following expression: $$\text{Area} = \frac{48r^2}{2\sin(\alpha)}-\frac{r^2}{2\sin(48\alpha)}$$ then a short computer program will confirm the maximum area as $366.5621$.

Thanks for editing my previous answer. It came out just as I had intended. Below is the program I wrote, using Power Basic.

cls:aold=0
 for r=14 to 16 step .001
  t=2*atn(1/sqr(4*r^2 - 1))
  anew=48*r^2/2*sin(t)-r^2/2*sin(48*t)
   if anew<aold then
     print (aold + anew)/2
     end
   end if
  aold = anew
 next

Answer 6

Sorry, I obviously have not mastered the proper formatting yet.The program written in Power Basic increases the value of r in increments of 0.001, and searches for a value where the area decreases. At that point the program computes the average value of the last two areas computed and prints that. Here again is the program.

cls:aold=0, 
for r=14 to 16 step 0.001,
t=2atan(1/sqr(4(r^2-1))
anew=48*r^2/2*sin(t)-r^2/2*sin(48*t)
if anew< aold then
print (aold+anew)/2,new line,end.
end if,new line,aold=anew
next