Find the area of largest rectangle that can be inscribed in an ellipse The actual problem reads: 

Find the area of the largest rectangle that can be inscribed in the ellipse 
  $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$

I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to find the derivative I don't think I'm doing it right. 
 A: $${1=\frac{{ x }^{ 2 }}{{ a }^{ 2 }}  + \frac {{ y }^{ 2 }} {{ b }^{ 2 }}}  \ge \frac{2 { xy }}{{ ab }} $$
when and only when  $${ x }/{ a } = { y }/{ b },$$  the max is got
i.e. :max of $xy =ab/2$, so $4xy=2ab$.
A: let L and H be the length and breadth of the required rectangle respectively
$$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$
$$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$
$$H=\frac{b}{a}\sqrt{4a^2-L^2}$$
Area=L*H
$$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$
$$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0$$
$$\frac{b*(4a^2-2L^2)}{a\sqrt{4a^2-L^2}}$$
$$=> 4a^2-2L^2=0$$
$$=2a^2=L^2$$
$$L=a\sqrt{2}$$
$$H=b\sqrt{2}$$
$$Area=L*H=2ab$$
$$\frac{d^2A}{dL^2}=\frac{\sqrt{4a^2-L^2}*(-4L)-\frac{4a^2-2L^2}{2\sqrt{4a^2-L^2}}}{4a^2-L^2}$$
Putting L=$a\sqrt{2}$
$$\frac{d^2A}{dL^2}=\frac{-a\sqrt{2}(4a\sqrt{2})-\frac{0}{2\sqrt{4a^2-2a^2}}}{4a^2-2a^2}$$
$$=\frac{-8a^2}{2a^2}$$
-4<0.
Therefore the area is max
A: Stretching the plane in a given direction is an operation that preserves ratios of areas. So:


*

*Stretch the plane by a factor of $a/b$ in the $y$-direction, to transform the ellipse to a circle with radius $a$.

*Inscribe the largest possible rectangle inside this circle, which turns out to be a square of area $2a^2$. Align this square with the $xy-$axes.

*Stretch the plane by a factor of $b/a$, to return the ellipse to its original size and shape. The resulting rectangle has area $2a^2\cdot b/a = 2ab$.
A: The vertices of any rectangle inscribed in an ellipse is given by $$(\pm a \cos(\theta), \pm b \sin(\theta))$$ The area of the rectangle is given by $$A(\theta) = 4ab \cos(\theta) \sin(\theta) = 2ab \sin(2 \theta)$$ Hence, the maximum is when $\sin(2 \theta) = 1$. Hence, the maximum area is when $2\theta = \dfrac{\pi}2$ i.e. $\theta = \dfrac{\pi}4$. The maximum area is $$A = 2ab$$
A: Suppose that the upper righthand corner of the rectangle is at the point $\langle x,y\rangle$. Then you know that the area of the rectangle is, as you say, $4xy$, and you know that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\;.\tag{1}$$
Thinking of the area as a function of $x$, we have $$\frac{dA}{dx}=4x\frac{dy}{dx}+4y\;.$$ Differentiating $(1)$ with respect to $x$, we have
$$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0\;,$$ so $$\frac{dy}{dx}=-\frac{b^2x}{a^2y}\;,$$ and $$\frac{dA}{dx}=4y-\frac{4b^2x^2}{a^2y}\;.$$
Setting this to $0$ and simplifying, we have $y^2=\dfrac{b^2x^2}{a^2}$. From $(1)$ we know that $$y^2=b^2-\frac{b^2x^2}{a^2}\;.$$ Thus, $y^2=b^2-y^2$, $2y^2=b^2$, and $\dfrac{y^2}{b^2}=\dfrac12$. Clearly, then, $\dfrac{x^2}{a^2}=\dfrac12$ as well, and the area is maximized when
$$x=\frac{a}{\sqrt2}=\frac{a\sqrt2}2\quad\text{and}\quad y=\frac{b}{\sqrt2}=\frac{b\sqrt2}2\;.$$
A: Parametric form of ellipse
$$ x = a \cos t,\; y = b \sin t, \; A = 4 x y = 2 a b \sin (2 t);$$
EDIT1:
Maximum area occurs for rectangle cutting by radial straight lines at $ t= \pm 45^0 $ through origin. The ellipse area is $\dfrac{2}{\pi}$ fraction of the enveloping rectangle area . The ellipse passes through rectangle corners $ \frac{a}{\sqrt2},\frac{b}{\sqrt2}.$ 
Due to two axis symmetry slanted orientations of rectangles can be ruled out.
A: Using the same technique shown here, we can orthogonally project the desired rectangle to the inscribed rectangle in the unit circle with maximal area (i.e. $2$, consider the inscribed square with sidelength $\sqrt{2}$). 
Let the maximal area of our rectangle be $\mathcal{A}$. Then, by preservation of area ratios, 
$$\begin{align*} \frac{\mathcal{A}}{\text{ Area of ellipse}} &=\frac{\text{Area of maximal square}}{\text{ Area of circle}}  \\\frac{\mathcal{A}}{ab\pi} &= \frac{2}{\pi}\\ \implies  \mathcal{A} &=2ab\end{align*}$$ which is our answer.
