Well, I eventually tried it again this morning, and got the answer. My method was completely right, strangely enough, doing what I was taught. I just picked the wrong variable to try and differentiate with. Doing it with what I had labeled as h is much easier than doing it with x or y.

This time I found the Area in terms of what I had labeled as <i>h</i>.

A=xy

Duh.

Well, x^2+h^2=16. Duh.

So x=(16-h^2)^(1/2).

Now I need to get y in terms of h. Obviously, y=h+4.

So my new equation that I used it this.

A=[(16-h^2)^(1/2)][h+4]

Now, find the derivative in terms of h, and then I'll set it to 0 to determine what h will be equal to at the maximum area.

dA/dh=(1/2)[(16-h^2)^(-1/2)](-2h)(h+4)+(1)(16-h^2)^(1/2)

This is really hard to type out.

OK, the (1/2) and the (-2h) can be combined, and just combining the first part of the equation gives you:

[(-h^2-4h)]/[(16-h^2)^(1/2)]+(16-h^2)^(1/2)=0

Now, I'm going to multiple everything by (16-h^2)^(1/2) to not only get rid of the radicals, but also get rid of the denominator in the first half of the equation.

[(-h^2-4h)]/[(16-h^2)^(1/2)] multipled by (16-h^2)^(1/2) just gives you (h^2-4h). Easy. (16-h^2)^(1/2) squared is just (16-h^2).

So the new equation is:

-h^2-4h + 16-h^2=0

-2h^2-4h+16=0

Factor out a -2

-2(h^2+2h-8)=0

The -2 can be discarded because it's a constant, and factoring the second part gets you:

(h+4)(h-2)=0

Obviously, h cannot equal -4, so h=2. Yay!

Doing the Pythagorean Theorem you get x^2+2^2=16, blah blah blah, x=2sqrt3. The entire base is 4sqrt3. The height is 6. And since Endless proved that it's an equilateral triangle, the / and \ are also 4sqrt3.

And I did all by myself this morning, too. Unfortunately I went to bed before I read any of you guys' help xD but thanks.