What is this supremum For any $10$ points in the unit circle, what is the value of the supremum of
the sum of the pairwise distances between the $10$ points, in which the supremum is taken over all configurations of 10 points?
 A: (take $\tau$$=2\pi$)
I feel like your question could be better explained, but it is an interesting question nonetheless (if I got it right). If the points can be anywhere without sequence, then five of them shall have coordinates such as $(\sin\frac\tau4;\cos\frac\tau4)=(0;1)$ and other five $(1;0)$, yielding a sum of $10$, 5 diameters, because the diameter is the biggest possible chord. In general, for $n$ points in the unit circle, the biggest sum of distances between them one can get is $2(n-1)$.
If they are in order (clock or counter-clockwise), then we can't jump like that, and in that case, if we stretch one chord too much, the others will be smaller. So the biggest will be the most similar to the circumference, a regular decagon. "I have a proof of this, which this page is to narrow to contain." :D Actually I don't, but am trying to find one (generally for all $n$, not just for the decagon, for which a proof could be found by brute force). Call it a conjecture for now.
In a regular decagon (inscribed in the unit circle), the angles from each point to the center (as compared to the positive part of the $Ox$ axis counter-clockwise) will be $+\frac\tau{10}$ each point, with the first at $0$ (for simplicity). Let's call the first point $A=(\sin0;\cos0)=(0;1)$. For an inscribed $n$-agon (having therefore $n$ points), each point in the biggest sum in counter-clockwise order shall be:
$$
P_i=\left(\sin\left(\frac{\tau i}n\right);\cos\left(\frac{\tau i}n\right)\right)
$$
It doesn't matter whether $i$ is the $0$-based index of points, until $n-1$, or a $1$-based index until $n$, as $\sin0=\sin\tau$, but as $A$ is the first point and on the $Ox$ axis, we'll have to go for the first. Note $\forall a\in \mathbb N_0, P_i=P_{i+an}$, because of the $\tau$ periodicity of the sine and cosine.
The distance between two points $A$ and $B$ is $\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$, by the Pythagorean theorem. So their sum will be:
$$
\sum_{i=0}^{n-1}
\sqrt{
\left(
\sin\left(\frac{\tau i}n\right)-\sin\left(\frac{\tau(i+1)}n\right)\right)^2
+\left(
\cos\left(\frac{\tau i}n\right)-\cos\left(\frac{\tau(i+1)}n\right)\right)^2
}=\\=
\sum_{i=0}^{n-1}
\sqrt{2
-2\sin\left(\frac{\tau i}n\right)\sin\left(\frac{\tau(i+1)}n\right)
-2\cos\left(\frac{\tau i}n\right)\cos\left(\frac{\tau(i+1)}n\right)}
=\\=
\sum_{i=0}^{n-1}\left(
\sqrt2\sqrt{1
-\sin\left(\frac{\tau i}n\right)\sin\left(\frac{\tau(i+1)}n\right)
-\cos\left(\frac{\tau i}n\right)\cos\left(\frac{\tau(i+1)}n\right)}\right)
=\\=
\sum_{i=0}^{n-1}\left(
\sqrt2\sqrt{1
+\cos\left(\frac{\tau(2i+1)}n\right)
-2\cos\left(\frac{\tau i}n\right)\cos\left(\frac{\tau(i+1)}n\right)
}\right)
$$
(note $P_0=P_n$, as generalized above; and remember the Pythagorean trigonometric identity plus some other properties I just learned about)
Applying to $n=10$ (thank God Wolfram Alpha understands $\LaTeX$, I just used 2*pi for \tau!) I got two different default exact results with the same approximation (I feel it works because the result is very close to $\tau=6,28...$. For those interested, I wasn't sure the trick I used to remove the sines worked so I tried both before and after. I digress away from the answer though, which is:
$$
5\sqrt{6-2\sqrt5}=\sqrt5+4\sqrt{6-2\sqrt5}-1=5\sqrt5-5\approx6,18
$$
I hope this is the answer you were looking for, I didn't use calculus though. I think it should be used (at least the concept of limits) to prove that conjecture, and I will add it later.
