I had been wondering while walking in corridors, me being lazy, what could be the shortest path between the two ends of the "L" shaped corridor.
Note: take the dimensions to be arbitrary.
Starting from any point from the 2cm line segment to any point in the 4cm line segment at the top.
Would going along the wall give the shortest path?
Or is there another path which is smaller than it?
This question has been solved please check out the sequel:
Optimised path 2.
Place the picture on a coordinate system with the lower left corner being the origin.
No matter which path we take, we have to go through a point of the form of $(4,t)$ where $t\in [0,2]$ from a point of the form of $(10,s)$ where $s\in [0,2]$, of which the distance is $\sqrt{(10-4)^2+(s-t)^2}\ge6$
The end point is $(p,9)$ where $p\in [0,4]$, to travel from $(4,t)$ to $(p,9)$, the distance is $$\sqrt{(4-p)^2+(9-t)^2}\ge\sqrt{(4-4)^2+(9-2)^2}=7$$
Hence a lower bound of the distance is $13$. We just have to walk along the inner wall, this holds for general L shape.