Minimum area of region bounded by a thread

Question

Suppose you have a thread of length $l$ and 4 thumbtacks located at $(1, 0), (0,1), (-1, 0), (0, -1)$. You have to make closed region with the thread so that all the 4 thumbtacks lie inside the region such that the area bounded by the thread is minimum. I'd like the area in terms of $l$. How would you go about this problem?

If $l \ge 8$, then area is $0$. And assume that $l\ge 4 \sqrt2$

Answer 1

We can simplify the problem using its symmetry, i.e., only consider the potion of the curve between $(0,1)$ and $(1,0)$.

Let $f$ be a continuous function defined from $[0,1]\to[0,1]$ such that

1. $$\int_0^1 \sqrt{1+(f'(x))^2}\textrm{dx} = \frac l4$$
2. $$\int_0^1 f(x) \textrm{ dx = minimum}$$

The first formula is obviously the standard length of a curve formula, and the second the area under the curve (which you will need to multiply by 4).

Unfortunately, the first formula is not really helpful. So we have to make some more clever observations.

By another symmetry argument, I argue that $f(x)$ reflected over the line $y=x$ will not change $f(x)$. Why should that be? Well, if the part of the function that reaches the line $y=x$ from $(1,0)$ is optimal, then the same sort of function should be used but reflected from $(0,1)$ to $y=x$.

The final clever observation to make is the following: the quarter circle is the optimal function to maximize area under the curve of a given length, so flipping that would minimize it.

While this is not a rigorous proof (Rigorous proofs do exist online), this should give you an intuitive understanding as to why curves of the form $$x^r+y^r=1$$ will be the correct function to minimize area.

You can now solve equation 1, but with $f(x) = \bigg(1-x^r\bigg)^\frac1r$ to find the correct $r$, and then use equation two to find the area under the curve