Consider the region bounded by the equations:
$x=1+2\pi$, $y=0$, and $(t+\cos(t),t+\sin(t))|0 \le t \le 2\pi$
I need to know how to find the coordinates of the green point $P$, which is arbitrary positioned for illustration, so that the sum of the distances from $P$ to the red closed curve is as small as possible.
See the following figure, for illustration;
Required to find the coordinates of $P$ such that $d_1+d_2+d_3+\dots+d_n$ is as small as possible.
I think, but not sure, that wherever $P$ is placed, the sum $d_1+d_2+d_3+\dots+d_n$ has no minimum value.
However, I believe that the standard deviation of $d_1,d_2,d_3,\dots,d_n$ can be minimized.
NOTE: $n \rightarrow \infty$. In other words, all points on the red curve are considered.
Providing me any hint or formula to be used would be appreciated. Thanks.
It seems like you're looking for the https://en.wikipedia.org/wiki/Geometric_median of the points of your curve. According to the Wikipedia page, there is a unique point that minimizes these distances for your shape (or any shape that isn't entirely along a line). Unfortunately, there is no simple formula for calculating these points. Various algorithms, including Weiszfeld's algorithm can provide an approximation.
I would encourage you to consider looking for the centroid of the points on your curve. In this case you would be minimizing the sum of the distances-squared from $P$ to your curve. This has a very nice formula: it's just the average of the coordinates of the points along your curve, and can be calculated with an integral, or approximated by computing an evenly distributed set of points along your curve and taking their average.
The centroid also more accurately reflects the real-world situation of trying to find the fairest location for locating a supermarket. Consider trying to locate a supermarket most fairly between two points $A$ and $B$. If you're trying to minimize the sum of the distances, any point on the line segment $AB$ works. If you're trying to minimize the sum of the squares of the distances, the midpoint does this uniquely.