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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

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    $\begingroup$ The general idea should work, though it is a little messy. You do not seem to have solved for $y$ in terms of $x$ correctly. Note that $y^2=12\left(1-\frac{x^2}{20}\right)$. $\endgroup$ – André Nicolas Nov 23 '15 at 3:38
  • $\begingroup$ Well, assuming your equation for $P'$ is correct, setting $P'=0$, moving the ugly fraction to one side, multiplying through by the denominator, and squaring both sides of the equation gives you a quadratic in $x$. And you know how to solve quadratics! EDIT: Just saw @AndréNicolas's comment - I think you should still get a quadratic? $\endgroup$ – Zubin Mukerjee Nov 23 '15 at 3:40
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    $\begingroup$ @ZubinMukerjee: Not only a quadratic, but one with no "$x$" term. $\endgroup$ – André Nicolas Nov 23 '15 at 4:12
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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 $4(x+y) = 4\sqrt{a^2+b^2}$.

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enter image description here

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 $$\bullet -\sqrt{a^2+b^2}\leq (a\cos \theta+b\sin\theta )\leq \sqrt{a^2+b^2}$$

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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

$$8\sqrt6\cdot\frac{\sqrt5}{\sqrt3}=8\sqrt10$$

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  • $\begingroup$ Why is it true that the rectangle that maximizes perimeter in the circle corresponds (by stretching in the $x$-direction) to the rectangle that maximizes perimeter in the ellipse? It's not obvious to me ... $\endgroup$ – Zubin Mukerjee Nov 23 '15 at 3:57
  • $\begingroup$ Well, it was true for area, but it doesn't seem to be true for perimeter based on Hamed's solution. Sorry about that. $\endgroup$ – Kay K. Nov 23 '15 at 4:01
  • $\begingroup$ Is it true for area? $\endgroup$ – Zubin Mukerjee Nov 23 '15 at 4:01
  • $\begingroup$ Yes. Otherwise, a square would have not been the maximum area inscribing rectangle inside a circle. $\endgroup$ – Kay K. Nov 23 '15 at 4:03
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    $\begingroup$ The stretching-into-a-circle is a cool way to do problems but you definitely have to be careful to note exactly why it is okay for you to reduce the problem to the circle version :) $\endgroup$ – Zubin Mukerjee Nov 23 '15 at 4:07
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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 $$p_{\max}=16\sqrt{2}\ .$$

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