<s>[INST] 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?
 [/INST] I agree with José Carlos Santos, yet I don't agree with your $A^2$ derivative. So, I like to show my way of solving with $A^2$.
$$Area =A= 4x \cdot \left(4 - \frac{4x^2}{9}\right)^{\frac12}$$
$$A^2= 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right) = 64x^2 - \frac{64x^4}{9}$$
Thus, derivative of $A^2$: $$ \frac1A\cdot\frac{dA}{dx} = 128x - \frac{256x^3}{9}=128x\left(1-\frac{2x^2}{9}\right)$$ Thus,
$$ \frac{dA}{dx} =A\cdot 128x\cdot \left(1-\frac{2x^2}{9}\right)$$
Since $A\ne 0$,  the critical values are: $x = 0$ or  $x=\pm \frac{3}{\sqrt{2}}$ because $\frac{dA}{dx}$ equals $0$ at critical points. Note that, as Professor Vector pointed out, "the rectangle with the largest area among those with sides parallel to the axes of the ellipse." There are two sides parallel to the $y$-axis, and the points of $x=+\frac{3}{\sqrt{2}}$ and $x=-\frac{3}{\sqrt{2}}$ are where these parallel sides meet the $x$-axis.
Consequently, the maximum area of rectangle is: $$A_{max}=$4\left(\frac{3}{\sqrt{2}}\right) \cdot \left(4 - \frac{4\left(\frac{3}{\sqrt{2}}\right)^2}{9}\right)^{\frac12}=12$$
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