I got this problem from Rustan Leino, who got it from Serdar Tasiran.
I solved it and wrote up my solution.
You're given one 44-meter piece of fence and 48 one-meter pieces of fence. Assume each piece is a straight and unbendable. What's the largest area of (flat) land that you can enclose using these fence pieces?
The answer is approximately $248.1~\mbox{m}^2,$ by the following reasoning.
Compute $r$ satisfying the equation $\arcsin(22/r) = 48 \arcsin(1/2r).$ One way to do this is via Wolfram Alpha using the substitution $x = 1/2r$; this gives $r \approx 33.5$.
Draw a circle with radius $r$ meters and set the 44-meter piece of fence inside it, with both points on the circle. Of those two points, one is more than 180 degrees clockwise from the other. Starting from that point, add the 48 one-meter pieces clockwise, always keeping the new pieces' endpoints on the circle.
Laying out the pieces in this way forms a closed polygon, by the following reasoning. Each endpoint of each fence piece is $r$ meters from the center of the circle. So, the 44-meter piece traces out an angle $2\arcsin(22/r)$ from that center, and each one-meter piece traces out an angle $2\arcsin(1/2r)$ from that center. This means that the 48-meter pieces collectively trace angle $96\arcsin(1/2r),$ which by the equation above is equal to $2\arcsin(22/r),$ the angle traced out by the 44-meter piece. Thus, the 48-piece arrangement abuts the 44-meter piece on both ends.
We've thus formed a polygon inscribed in a circle. By the following lemma, this means that it has maximum possible area among polygons that can be formed by the pieces given.
Lemma. Suppose two polygons $P_1$ and $P_2$ have the same side lengths, and $P_1$ inscribed in a circle. $P_1$'s area is at least that of $P_2.$
Proof. This proof is due to David Eppstein, and summarized at the Hall of Hexagons website.
Without loss of generality, we can assume $P_2$ is convex, since if it isn't we can make it even bigger by "popping out" its concave sides. Consider the part of the circle $P_1$ is inscribed in that lie outside $P_1.$ This can be divided into arc-shaped pieces, each of which abuts one of the sides of $P_1.$ We can make copies of these pieces and "glue" them onto the corresponding same-length sides of $P_2.$ In this way, we construct a new shape with the same perimeter as the circle. Since circles have maximum area among shapes with the same perimeter, this new shape's area must be no more than that of the circle. But, this new shape's area is the total area of the arc-shaped pieces plus the area of $P_2,$ and the circle's area is the total area of the arc-shaped pieces plus the area of $P_1.$ Thus, $P_2$'s area is at most that of $P_1.$
We can compute the area of the polygon created this way as follows. For each side, let its $\cal A$ value be the area of the triangle formed by connecting the endpoints of the side to the center of the circle. The area of our polygon is the sum of the 48 $\cal A$ values for the one-meter sides, minus the area of the $\cal A$ value for the 44-meter side.
The $\cal A$ value for a side of length $s$ meters is readily computed since it's the area of an isoceles triangle. It's $\frac{s}{2}\sqrt{r^2 - \left(\frac{s}{2}\right)^2}$ meters. So, the area of our polygon is $$ 48 \cdot \frac{1}{2} \sqrt{r^2 - \left(\frac{1}{2}\right)^2} - \frac{44}{2} \sqrt{r^2 - \left(\frac{44}{2}\right)^2}$$ meters. Computing this value gives $\approx 248.1~\mbox{m}^2.$