Differentiated values are treated just like any other variable, right?

[Juxtaroberto]

http://www.algebra.com/algebra/home...aneous_Word_Problems.faq.question.596257.html

One of the steps leads us to

A=2x√(r²-x²).

The page then says that we could differentiate at this step, but that it would be easier to first square both sides, so as to not have to apply the product rule and the chain rule, and we end up with

A²=4x²(r²-x²)
A²=4r²x²-4x⁴

Then we differentiate both sides with respect to x

2A dA/dx=8r²x-16x³

This is where it gets weird, and I don't know if I am simply not advanced enough in math, or if this person made a mistake. He then divides both sides by 2A to isolate dA/dx, but after he does it, he ends up with

dA/dx=(8r²x-16x³)/A

What happened to the 2? It wasn't reduced out of the numerator... can we just drop it? Furthermore, in a later step, he multiplies both sides of the equation by A... I thought we weren't generally allowed to do that, because it removes possible solutions? I'm assuming this time it was allowed because A stands for the area in general, and not the variable x we were trying to solve... right?

I hope someone can help me with this...

 

[pwsnafu]

Yes you are right: he forgot to divide by two. It doesn't affect his answer in the end, because he is setting dA/dx = 0. But yes, he would lose marks. As for your second question, we are only interested in rectangles with strictly positive area, ##A\neq 0##. It's the context of the question.

As an aside, this is an example of when you shouldn't use calculus. From the generalized mean inequality we have
[itex]A = 2xy = 2 M_0(x,y)^2 \leq 2 M_2(x,y)^2 = 2 \frac{r^2}{2} = r^2[/itex] with equality when x = y. Yeah, it's a one liner.