Given that we cut a corner, the longest possible value for the original problem is √2.
If we want the proportion cut, we get (1 - 1/√2) ≅ 0.3
The important realisation that the ratio of cutting corners that is maximised at 45° is not shown by the graph, which is a shame.
You don't have to use trig to show that, but I think it helps.
Edit: as a side point, as I was thinking about this, I realised that you've discovered the maximum relative error of an L1 norm vs an L2 norm, which is fun
The x-axis in my plot shows the ratio of lengths of the longer to the shorter of the two non-hypotenuse sides (yes, typing this out it does sound convoluted). This ratio is 1 when those two sides are equal length, which implies that the two corners are 45°. So the plot does show that the return to cutting corners is maximized at 45°-45°-90°, but indirectly.
It's not an arbitrary decision that the graph only covers half the angle range. That's a consequence of the decision to label the shortest side "a" and the longer one (non-hypotenuse) "b". If those labels were applied to specific sides then you could explore the full angle range, but it would be redundant. I'm open to the idea that people might find this easier to understand though.
I think the natural way to refer to the sides of a triangle is by their length, which lead to my formulation. If you feel like plotting your formulation, here's my source code (in R): https://gist.github.com/masonicboom/dafeb49b3c1d4c44998969ef....
Cheers!