$12x^2-12x = 0$
This was suppose to be 18, but it is -18.
I think it is because of how I modified one of the factors.
$\sqrt{x^2+x^2}$
$\sqrt{2x^2}$
$\sqrt{2}\cdot \sqrt{x^2}$
$x\sqrt{2}$
If I let wolfram do this, it gives the same but it says "assuming X is positive". Isn't that redundant? Any negative number squared is positive? The problem could be with the other factor that I modified.
$\sqrt{(6-x)^2+(6-x)^2}$
$x\sqrt2 -6\sqrt2$
To sum it up:
$x\sqrt2(x\sqrt2-6\sqrt2) = 2x^2-12x$
It should be $-2x^2+12x$, because that would indeed give positive 18 as it's answer.
Update! With all the info!
The task is to find x such that the area of the rectangle would be as big as possible. I used the Pythagorean theorem and wrote out the following equation.
$\sqrt{(6-x)^2+(6-x)^2} \cdot \sqrt{x^2+x^2} = A$ of the rectangle.
The left factor would be the long side of the rectangle and the right factor would be the short side. I need to write this out nicely and then find when the derivative is 0. For max area.