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!
 A: Let $\theta$ be the angle that the slanted red (?) line on the right makes with the horizontal.
Then the height of the rectangle is $r\sin\theta$ and the base is $2r\cos\theta$, for an area of $r^2\sin\theta\cos\theta$.  
This is $\frac{r^2}{2}\sin 2\theta$. But $\sin 2\theta$ has a maximum value of $1$, at $\theta=\frac{\pi}{4}$.
A: You have dropped an $x$ in calculating your derivative.  By applying the product rule:
$$\begin{align}A'(x) &= 2x\left(\frac{1}{2}(r^2-x^2)^{-1/2}(-2\color{red}{x})\right) + 2\sqrt{r^2-x^2}\\
&= \frac{-2x^{\color{red}{2}}}{\sqrt{r^2-x^2}} + 2\sqrt{r^2-x^2}\end{align}$$
A: hint :$x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2}$
A: 
Eqn of circle: x^2 + y^2 = r^2
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area of rectangle = a = 2xy
==> a^2 = 4 x^2 y^2
==> a^2 = 4 x^2 (r^2 - x^2)
==> a^2 = 4 (x^2 r^2 - x^4)
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differentiating:
==> 2a da/dt = 4 (2x r^2 - 4x^3)
since, da/dt=0
==> 2 x r^2 = 4 x^3
==> x = r/sqrt(2)
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second derivative test:
==>4 (2 r^2 - 12 x^2)
==>8 r^2 - 24 r^2
==>(-16) r^2
==> -ve
therefore, maximum area for  x= r/sqrt(2)
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finding y:
==>y^2 = r^2 - x^2
==>y^2 = r^2 - (r^2)/2
==>y=r/sqrt(2)
A: 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)
A: By elementary means:
The maximum of $2x\sqrt{r^2-x^2}$ is also that of $x^2\sqrt{r^2-x^2}$, by squaring, or that of $t(1-t)$, by setting $r^2t=x^2$.
The expression $t(1-t)$ describes a parabola, with roots at $t=0$ and $t=1$, and by symmetry its vertex must be at $t=\dfrac12$.
Hence the requested area,
$$2\frac r{\sqrt 2}\sqrt{r^2-\frac{r^2}2}=r^2.$$
A: Doubling the semi-circle to obtain a full circle, we now have a rectangle inscribed in a circle, with area equal to twice the area of your starting rectangle. Letting $x$ be either angle formed by its diagonals, we need to maximize
$$
A(x)=2r \sin x\,.
$$
This has a maximum when the diagonals are orthogonal, i.e. when the rectangle is a square, with side $\sqrt 2\, r$. Hence, the dimensions of the original rectangle are $\sqrt2\,r$ and $r/\sqrt2\,$.
A: 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, $\mathbf{max. A = \sqrt{r^4} = r^2}$
