maximum area of a rectangle inscribed in a semi - circle with radius r. 
A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum. 

My Try: 
Let length of the side be $x$,
Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.

Then the area function is 
$$A(x) = 2x\sqrt{r^2-x^2}$$
$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\
&=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$
setting $A'(x) = 0$,
$$\implies x^2 +2x -r^2 = 0$$
Solving, I obtained:
$$x = -1 \pm \sqrt{1+r^2}$$
That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation. 
Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious. 
Thanks!

Equation of circle: $x^2+y^2=r^2$
$x = \pm (r^2-y^2) $
Thus, length of rectangle is $x-(-x)=2x$ and height is $y$.
Area $A = 2xy$. Maximizing A is equivalent to maximizing $ A^2$
$A^2 = 4x^2y^2 = 4(r^2-y^2)y^2=4r^2y^2-4y^4$
Let, $f(z)= 4r^2z-4z^2 $
Then, $f'(z) = 4r^2-8z$
Equating, $f'(z)=0$ we get,
$\mathbf{z=\frac12r^2}$
Therefore, max. $A^2$ or max. $f(z) = 4r^2\times \frac12r^2-4\times(\frac12r^2)^2 = 2r^4-r^4=r^4$
Hence, $\mathbf{max. A = \sqrt{r^4} = r^2}$
