Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? 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:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$
$$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$
Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?
So this track seems to difficult, another approach. Can I square the area first, find the derivative of that to solve for x?
So the Area = $4x \cdot \sqrt{4 - \frac{4x^2}{9}}$
Is this valid?
$$Area^2 = 16x^2 \cdot (4 - \frac{4x^2}{9}$$
$$= 64x^2 - \frac{64x^4}{9}$$
Derivative: 
$$ \frac{d}{dx} Area^2 = 128x - \frac{256x^3}{9}$$
$$128x(1-\frac{2x^2}{9}$$
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 Area is 3.
Is this valid? If so why? Does squaring not cause any problems?
 A: You are over-complicating it. Affine maps preserve the ratios of areas and, in a circle, the inscribed squares are pretty obviously the largest inscribed quadrilaterals. Their area is $\frac{2}{\pi}$ times the area of the circle. By applying $\varphi:(x,y)\mapsto(ax,by)$ the ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is mapped into a circle. By applying $\varphi^{-1}$ we get that

A. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the largest inscribed rectangle (which is also one of the largest inscribed quadrilaterals) is symmetric with respect to the axis of the ellipse and its area equals $2ab$.

By the very same principle,

B. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the area of the largest inscribed triangle equals $\frac{3\sqrt{3}}{4}ab$.
C. In a triangle $ABC$ the largest inscribed ellipse is tangent to the sides at their midpoints and the center of such ellipse lies at the centroid of $ABC$. This is the Steiner inellipse of $ABC$.

A: Hint...it would be a lot easier if you wrote $$(x,y)=(3\cos\theta,2\sin\theta)$$ then obtain an expression for the area of the rectangle in terms of $\theta$ and differentiate...
A: I think you are in right track but your expression for the first derivative of $A$ need a little touch. I put it in more clear way:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}=4x \left(4-\frac{4x^2}{9}\right)^{1/2}$$ Thus,
$$A' = 4 \left(4-\frac{4x^2}{9}\right)^{1/2} + 4x \left(4-\frac{4x^2}{9}\right)^{-1/2}\left(\frac{-8x}{9} \right)=4 \left(4-\frac{4x^2}{9}\right)^{1/2} - \frac{32x^2}{9} \left(4-\frac{4x^2}{9}\right)^{-1/2}=\frac{36 \left(4-\frac{4x^2}{9}\right) - 32x^2} {9\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{4 \left(36-4x^2\right) - 288x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{144 - 304x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}$$
To find the maximum for $A$, set $A'=0$ since $-3\le x\le +3$, and hence $\left(4-\frac{4x^2}{9}\right)^{1/2}\ne0$. Therefore, $144 - 304x^2=0$ when $A'-0$ and $x=\pm\sqrt\frac{19}{9}$.
$$A_{max}=4\left(\sqrt\frac{19}{9}\right) \left( \sqrt{4 - \frac{4\left(\sqrt\frac{19}{9}\right)^2}{9}}\right)=8\left(\sqrt\frac{19}{9}\right) \left( \frac{\sqrt{62}}{9}\right)=\frac{8\sqrt{1178}}{27}$$
Note: The $\pm x$ values for rectangle are symmetrical points on major axis of ellipse where vertical legs of rectangle meet.
