Distances over a rectangular parallelepiped It is given a rectangular parallelepiped $3\times4\times5$. Which are the farthest points from a given vertex, provided one can only walk on the surface of the parallelepiped?
[Edit, for clarification]
The farthest point is not the opposite vertex, which has a distance equal to $\sqrt{74}\approx8.6$, as one can see looking at the following picture:

If one take a point $R$ which lies $0.3$ to the left of $F$ and $0.1$ above $F$, the distance $AR$ is equal to $\sqrt{\dfrac{757}{10}}\approx8.7$, greater than $AF$.
Besides, the other path along the $4\times5$ and the $3\times4$ faces, which leads to point $S$, is even longer.
I think that the point that I'm searching for should have the property that $AS=AR$ (and, of course, that $AR$ should be maximal).
 A: The farthest spatial point from one vertex is at the other end of the space diagonal which has a distance of $\sqrt{3^2+4^2+5^2}=\sqrt{50}$.
There's only shortest path along the surface of a cuboid among these two points.  By drawing a net of the cuboid, you can see how to find it out.

The shortest path is the minimum among the following:
$$\sqrt{a^2+b^2+c^2+2bc} \, , \; \sqrt{a^2+b^2+c^2+2ca} \, , \; \sqrt{a^2+b^2+c^2+2ab}$$
That is $\sqrt{5^2+4^2+3^2+2(3)(4)}=\sqrt{74}$
By the way, the spatial distance never exceeds the surface distance.
A: There are 4 different ways to get from the corner to a point in the opposite small face, as shown in the picture below. Note that the way described by AN, going below the cube, is a shorter way to get to your R/S point in the question, and that point is actually worse than the corner.

For an AxBxC cuboid, if we denote the starting corner as (0,0,0), the opposite corner as (C,B,A), and our target point as (C, B-x, A-y), we have the distances:
$AG = (C + A - x)^2 + (B - y)^2 \\
AJ = (B + A - x)^2 + (C + y)^2 \\
AV = (A - x)^2 + (C + B - y)^2 \\
AN = (C + x)^2 + (A + B - y)^2$
Therefore our optimal point is either the opposite corner, or a point on one of the lines determined by two of the above distances being equal. I recommend you check all lines by yourself to check my answer, but I found that for the specific case of the 3x4x5 cuboid the optimal point is, indeed, the opposite corner.
