Maximum area of 3 rectangles inside an ellipse I'm trying to determine the maximum area in a specific ellipse  that can be filled with any 3 (horizontally aligned) rectangles. 
$$Ellipse: \frac{x^2}{36}+\frac{y^2}{16}=1$$
$$Area: A=4(x_1*y_1+(x_2-x_1)*y_2)$$
Here's an image:

By substituting $y_1$ and $y_2$ with a function of $x_1$ and $x_2$ (with the equation of the ellipse), differentiating that and equal it to 0, I get $x_1$ as a function of $x_2$. I don't think that's the solution, so how can I solve this? Is there even a finite number of solutions?
Hope you can/will help me :)
 A: (It seems that I do not have privilege to post this partial solution as a comment to your problem, so I write it here. Wish it can help you)
First  I assume that all of the vertices of the middle rectangle touch the ellipse, that is, the middle rectangle is symmetrical along the $X$ axis.
Secondly,  I set $x1$ $x2$ $x3$ as bellow:

Then you need to maximize the area:
$S=2*(x1-x2)*y1 + 4*x2*y2 + 2(-x2-x3)*y3$
theoretically you can use Lagrange Multiplier to solve this problem, but it has too many variables for computing by hand.
So I use Mathematica here:

    NMaximize[
 {2 (x1 - x2)*y1 + 4*x2*y2 - 
   2 (x2 + x3)*y3, (x1^2)/36 + (y1^2)/16 - 1 == 
    0 && (x2^2)/36 + (y2^2)/16 - 1 == 
    0 && (x3^2)/36 + (y3^2)/16 - 1 == 0 && y1 > 0 && y2 > 0 && 
   y3 > 0}, {x1, x2, x3, y1, y2, y3}]

It gives that

    {59.3313, {x1 -> 5.10391, x2 -> 3.15439, x3 -> -5.1039, y1 -> 2.10292,
   y2 -> 3.4026, y3 -> 2.10292}}

Above solution bases on the assumption that the middle rectangle is symmetrical along the $x$ axis. So it still needs some discuss on the non-symmetrical situation.
A: The simpler solution for me is the following one:
First, note that this problem is about area, which changes linearly with stretching, so you can solve the problem in the unit circle ($x^2+y^2=1$), and then find the solution by multiplying the $x$ by $6$ and the $y$ by $4$.
Next, note that the solution doesn't change if the biggest rectangle is horizontal or vertical, so the solution should not change if you swap $x$ and $y$ (otherwise, this can't be a maximum), so $y_1=\sqrt{1-x_1}$ should be equal to $x_2$, so we have to fix only $x_1$, and then $x_2=\sqrt{1-x_1}$ and $y_2=x_1$.
Now you have an equation of a single variable:
$f(x)=4\left(x\sqrt{1-x^2}+\left(\sqrt{1-x^2}-x\right)x\right)$,
so you can find the derivative, and then its root, which is
$\sqrt{\frac{1}{10}\left(5 - \sqrt{5}\right)} = 0.525731 \dots$
This gives a maximal area of $4\cdot 6\cdot f(x)=59.3312\dots$ for $x_1=x\cdot6=3.15439$ and $x_2=\sqrt{1-x^2}\cdot6=5.1039$
(the same result as Henry and craftsman.don).
