Enclosing land by fence pieces 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?

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:
$$B \approx 30.562$$
$$A \approx 366.562$$ 
