Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:
$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$
$$y^2 = 4 - \frac{4x^2}{9}$$
$$y = \sqrt{4 - \frac{4x^2}{9}}$$
So the area function is now:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?
So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative: 
$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$
$$128x\left(1-\frac{2x^2}{9}\right)$$
So critical values: $x = 0, \frac{3}{\sqrt{2}}$
because the derivative equals $0$ when:
$$2x^2 = 9$$
$$x = \frac{3}{\sqrt{2}}$$
Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.
Is this valid? If so why? Does squaring not cause any problems?

Starting from
$$
\left\{ \matrix{
  {{x^{\,2} } \over 9} + {{y^{\,2} } \over 4} = 1 \hfill \cr 
  A = 4xy \hfill \cr}  \right.
$$
which gives the intercepts between the ellipse and a hyperbola,
you could have simply substitute $y=A/(4x)$, to get
$$
{{x^{\,2} } \over 9} + {{A^{\,2} } \over {64x^{\,2} }} = 1
$$
i.e.
$$
64x^{\,4}  - 9 \cdot 64x^{\,2}  + 9A^{\,2}  = 0
$$
put the discriminant to be null, and get $A$
$$
\Delta  = \left( {9 \cdot 64} \right)^{\,2}  - 4 \cdot 64 \cdot 9A^{\,2}  = 0\quad  \Rightarrow \quad A^{\,2}  = 9 \cdot 16\quad  \Rightarrow \quad A = 12
$$
and since the discriminant is null, from the above it follows
$$
x^{\,2}  = {{9 \cdot 64} \over {2 \cdot 64}}\quad  \Rightarrow \quad x = {3 \over 2}\sqrt 2 \quad  \Rightarrow \quad y = A/4x = \sqrt 2 
$$
