# The Center of Math Blog

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## Monday, June 15, 2015

### Problem of the Week

Good morning from the Center of Math! This problem today seems like a geometry problem when you first look at it, but I solved it with calculus. Take a look at the question below, which I posted to our Facebook, Twitter, and Google+ pages.

 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!

-Tori

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):

The problem is the same as to find the right triangle with the greatest area among those of hypotenuse L. These are all inscribed in a semi circle of diameter AB=L, our best guess is that the isosceles triangle ABC has the greatest area, i.e. greater than any triangle ABD. This is the same as to prove that the area of the blue triangle BCI is greater than the area of the red ADI. The blue and red triangles have the same angles but the hypotenuse of the blue is greater than that of the red, indeed:

BI>BC=AC>AI.

Hence the greatest area is achieved with the isosceles triangle ABC whose area is L²/4 (one quarter of the square of side L)

1. The answer for a given L is indeed L²/4 and here is a geometrical proof without calculus:
I can't post a figure here, so I'll let you draw it. (I posted one on Facebook, though)
The problem is the same as to find the right triangle with the greatest area among those of hypotenuse L. These are all inscribed in a semi circle of diameter AB=L, our best guess is that the isosceles triangle ABC has the greatest area, i.e. greater than any triangle ABD with D on the semi-circle. Say that AC intercepts BD in I. We want to prove that the area of the triangle BCI is greater than the area of ADI. These triangles have the same angles but the hypotenuse of BCI is greater than that of ADI, indeed:
BI>BC=AC>AI.
So the area of BCI is greater than that of ADI.
Hence the greatest area is achieved with the isosceles triangle ABC whose area is L²/4 (one quarter of the square of side L)

Vincent Pantaloni (Orléans, France)

2. Sorry, but my English might not be good enough. I don't understan 'the maximum area of the resulting triangle'. A triangle has just one area, not a maximum and a minimum one. You found the area for the triangle with the biggest area possible for the given conditions, which of course is going to be when L gets to 1.4142.... what means almost L(sq2)/4 = 2/4 =0.5. If it isn't like that, I think I'll have to keep learning more English! :/

3. I think "maximum" isn't quite the right word, since if your line is strictly shorter than sqrt(2) in length, you will never be able to place it such that you get the triangle of area 1/2

What that is, is the supremum. For any value A < 1/2, there exists some L < sqrt(2) such that the triangle formed from the line of length L can have area A.

So as A gets arbitrarily close to 1/2 (and it can get as close as you'd like, as long as it doesn't reach 1/2), then you can get some L < sqrt(2) to form the desired triangle. The goal of A = 1/2 is never possible, but as you can get any value smaller than A = 1/2, this is the "supremum" of possible areas, the least upper bound on the possible areas.