This problem was posted by our own Agnishom in another thread.
A triangle has sides of length at most 2, 3 & 4. What is the maximum area the triangle can have?
Let's see if geogebra 4.27 can lend some insight into the problem.
1) Hide the xy axes.
2) Place point A anywhere on the screen.
3) Use the circle with center and radius tool and click on point A and enter a radius of 2.
4) Use the point on object tool to create a point B on the circle's circumference. See the first drawing.
5) Use the circle with center and radius tool and click on point B and enter a radius of 3. A second circle will be created.
6) Use the point on object tool to create a point C on the larger circle's circumference. See the second drawing.
7) Use the polygon tool and click A, B, C, and back to A. A triangle will be created with sides AB = 2 and BC = 3. Move the three vertices around to see that those sides are constant. See fig 3.
You can see on the left the brown algebra pane which gived the sides and the areas of that triangle.
8) Set rounding to 15 decimal places in options. Play with points B and C carefully with your mouse or shift right, left arrows and try get the largest area for poly1 while keeping b < = 4.
It is not too difficult to reach poly1 = 2.999992275497143. if you are lucky you will even get poly1 = 3. No matter how hard you try you will not get an area bigger than 3. Once you are satisfied with your value use the angle tool to measure angle ABC. I got α = 89.86997934032091°. Hence the assumption of 90°, which yields sides of 2,3, √(13), with a maximum area of 3.
What has been accomplished here? For one thing we have a conjecture for the 3rd side and the conjecture that this has to be a right triangle. This helps in looking in the literature for a possible answer based on that. Or might guide us in looking for an answer.
If we are are unable to find the maximum we at least have a good estimate.