How do I find the maximum perimeter of a rectangle inscribed in an ellipse? The problem I've been stuck on is this:
A rectangle is inscribed in the ellipse  $$\frac{x^2}{20} + \frac{y^2}{12} = 1$$ What is the maximum perimeter of the rectangle?
I don't even know if I'm taking the right approach. So far, I've been trying to solve for $y$, giving me $y = \sqrt{12-(3/5)x}$, and plugging that into the equation $P = 4x + 4y$, which should be the equation for the perimeter of an inscribed rectangle. I then took the derivative of $P$ after plugging in the equation for $y$, giving me $$P' = 4 - \frac{12x}{5\sqrt{12-(3/5)x}}.$$ To find a maximum, I'd set the equation to zero right? Well, I don't know where to go from this step, since simplifying from here only seems to make it harder.
Any help would be much appreciated, even a nudge in the right direction. I have no idea where to go from here, or even if I got to the right place. Thanks for your time
 A: One simple way of solving this problem is by Lagrange multipliers method. Note that if $(x,y)$ is in the first quadrant on the ellipse $x^2/a^2+y^2/b^2 = 1$, then the perimeter of the inscribed rectangle represented by $(x,y)$ is simply $4(x+y)$. Therefore you want to maximize $x+y$ given the constraint that $x^2/a^2+y^2/b^2 = 1$. Define
$$
f(x,y,\lambda) = x+y -\lambda\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)
$$
Hence by maximizing $f$
$$
1 = \frac{2x\lambda}{a^2}=\frac{2y\lambda}{b^2}\Longrightarrow \frac{x}{a} = \frac{y}{b}\left(\frac{a}{b}\right)
$$
but then
$$1=\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{y^2}{b^2}\left(1+\frac{a^2}{b^2}\right)\Longrightarrow y=\frac{b^2}{\sqrt{a^2+b^2}}, \quad x=\frac{a^2}{\sqrt{a^2+b^2}}$$
The maximum perimeter is therefore $4(x+y) = 4\sqrt{a^2+b^2}$.
A: 
Let equation of ellipse be $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\;,$ Then we will take variable point $P,Q,R,S$ 
on that ellipse, and parametric Coordinate of  Point $P(a\cos \theta,b\sin \theta).$
Similarly $Q(-a \cos \theta,b\sin \theta)$ and $R(-a \cos \theta,-b\sin \theta)$ and $S(a \cos \theta,-b\sin \theta)$
So Paramteter of Recatangle is $$\displaystyle P=4a\cos \theta+4b\sin \theta =4(a\cos \theta+b\sin \theta)\leq 4\sqrt{a^2+b^2}.$$
Above we have used the formula $$\bullet -\sqrt{a^2+b^2}\leq (a\cos \theta+b\sin\theta )\leq \sqrt{a^2+b^2}$$
A: Let me squeeze the ellipse into a circle:
$$\frac{x^2}{12}+\frac{y^2}{12}=1$$
And I would claim that the maximum perimeter rectangle inside the circle is the square. Its perimeter is
$$4\sqrt2\space r = 4\sqrt2\cdot2\sqrt3 = 8\sqrt6$$
Now let me recover the circle back to an ellipse. And the square is also stretched into a rectangle and one of its side is magnified by factor of $\sqrt{20/12}=\sqrt{5/3}$. And then the new perimeter is
$$8\sqrt6\cdot\frac{\sqrt5}{\sqrt3}=8\sqrt10$$
A: All rectangles $[{-a},a]\times[{-b},b]$ with given perimeter $p$ have the vertex $P=(a,b)$ on the line $$\ell_p:\quad a+b={p\over4}$$ of slope $-1$. Increasing $p$ means that $\ell_p$ is translated north-east. The largest $p$ that can be realized for a $P$ on the given ellipse $$E:\qquad f(x,y):=3x^2+5y^2=60\tag{1}$$ is when $\ell_p$ is tangent to $E$. We therefore have to find the point on $E$ in the first quadrant where
 $\nabla f(x,y)=(6x,10y)$ points due north-east. This enforces $y={3\over5}x$, so that we obtain from $(1)$ the point $P={1\over\sqrt{2}}(5,3)$, leading to the maximal perimeter $$p_{\max}=16\sqrt{2}\ .$$
