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!

For convenience, think that the circle has its center at (0,0). We then consider the upper semicircle of x^2+y^2=a^2. (1)
The area of the inscribed rectangle would be A=2xy
dA/dx=(2x)'y+2x(dy/dx)
=2y+2x(dy/dx)
diff (1)
(d/dx)(x^2+y^2)=(d/dx)(a^2) 
<=> 2x+2y(dy/dx)=0 
<=> dy/dx = -x/y
Now, dA/dx = 2y+2x(-x/y) 
<=> dA/dx = 2(y^2-x^2)/y
<=>(1) dA/dx = 2(a^2-2x^2)/y
dA/dx = 0
<=> a^2-2x^2 = 0
<=> x = a/sqrt(2)
