If $y = cx^2$ for $c>0$ and $0 \leq x \leq 1$, what is the largest possible area of the rectangle? Consider the collection of rectangles with a vertex at $(1,0)$ and the other lying on $ y = cx^2 $ for some $c > 0$ and $0 \leq x \leq 1$. Find the rectangle with maximal area.
So I named the height as $y$ and width as $x$ and the area formula became $ A = xy$. Then I use the given equation $y=cx^2$ to sub in y in the area formula and I got $A = x\cdot cx^2$ which became $A = cx^3$. Then I took the derivative to find the max area and I got $A' = 3cx^2$. When I tried to make it equal to $0$, only critical point is $x = 0$ and it doesn't make sense since a rectangle's width cannot be $0$.
I don't know where I am making a mistake, so please help me. Thank you in advance.

I assume that sides of the rectangle are parallel to axis. Then the desired area is:
$$ A(x)=cx^2|x-1|=cx^2(1-x)$$ The equality goes from the fact that $x\in(0,1)$.
Then taking the derivative 
$$A'(x)=2cx(1-x)-cx^2$$
The derivative is zero iff
$$2cx(1-x)-cx^2=cx(2-3x)=0$$
Then for $x=\frac{2}{3}$ the area is maximal (you can take the second derivative to see that it is indeed a maximum).
EDIT: To your solution, you mistakenly took the length of the $x$-axis side to be $x$, but it is $|x-1|$.
