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!

hint :$x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2}$
