|You can click on any photo in the post to expand|
And as every week these past six months or so, I've written down my solution. Take a look below...
Above, I drew out my square again and made sure I recognized what the question was asking for.
And then I drew a couple different squares with arbitrary lines L, and looked at how different triangles can be formed. Obviously, the largest possible triangle that can be made occurs when L lies directly across the square, cutting it in half. Using the Pythagorean theorem, I found out that L is equal to the square root of 2 here, and our area A is equal to 1/2. Makes sense, this is half of the Unit Square.
But our question asks for when L is less than the square root of 2 in length! So, I had to do some algebra to isolate a variable or two. I found an equation for Area (A) that uses just one variable (x) for any given L.
I used calculus to solve the problem from here. I found the derivative of the area equation, and used the methods of the first derivative test to find the value of x for a maximum A (continues into the next image).
After I found the x value that maximizes A, I plugged it back into the equation for Area. After some algebra, we've found a general equation that gives us the maximum area for any length L.
If you have any questions about how I solved this problem, let me know in the comments!
UPDATE: One of our followers (and blog commenters) sent in a geometrical proof that doesn't use calculus. He actually did this on Monday, but I finally have time to edit my post here. Thank you Vincent Pantaloni! Here's his image:
And the process he supplied, as stated on the Center of Math Facebook post (it's also in our comment section here):
Hence the greatest area is achieved with the isosceles triangle ABC whose area is L²/4 (one quarter of the square of side L)