How to find the maximum perimeter of rectangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Question

Find the maximum perimeter of a rectangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

I was able to do until $P=4(x+y)$.

$$\frac{\mathrm dP}{\mathrm dx} =4-\frac{4b}{a^2}\left(1-\frac{x^2}{a^2}\right)^{\frac12}$$

But can't figure how to solve $\frac{\mathrm dP}{\mathrm dx} =0$.

Could someone help me out by showing me step by step how to solve it?

Answer 1

This is an interesting problem, so I'm going to go through the solution. We can uniquely determine a rectangle inside an ellipse given exactly one point. The other three points simply follow from finding the intersections of the ellipse with some straight lines constructed through the given point. See here for a visualization of this. Assume that we pick this point from the first quadrant and construct the other three. So our optimization problem is as follows:

Maximize $P(x,y)=4(x+y)$ under the constraints $f(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ and $a,b>0$; $x\in[0,a]~;~y\in[0,b]$

Seems like a perfect time to use the method of Lagrange multipliers. Some computations: $$\nabla P=(4,4)$$ $$\nabla f(x,y)=2\left(\frac{x}{a^2},\frac{y}{b^2}\right)$$ We need to find points $(x,y)$ such that $\nabla f=\lambda \nabla P$ for some $\lambda\in\Bbb{R}$. So we have a system of equations: $$4\lambda=\frac{2x}{a^2}~;~ 4\lambda=\frac{2y}{b^2}~;~ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Substituting $y=b\sqrt{1-\frac{x^2}{a^2}}$ we can reduce to a two variable system: $$4\lambda=\frac{2x}{a^2}~;~4\lambda=\frac{2\sqrt{1-\frac{x^2}{a^2}}}{b}$$ So now we need to solve the equation $$\frac{x}{a^2}=\frac{1}{b}\sqrt{1-\frac{x^2}{a^2}}$$ Squaring, $$\frac{x^2}{a^4}=\frac{1}{b^2}\left(1-\frac{x^2}{a^2}\right)$$ $$\frac{x^2}{a^2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)=\frac{1}{b^2}$$ $$x=\frac{a^2}{\sqrt{a^2+b^2}}$$ Similarly $$y=\frac{b^2}{\sqrt{a^2+b^2}}$$ Thus our maximal perimeter is $$P_\max=4\sqrt{a^2+b^2}.$$

And an alternative approach. We can also parameterize the first quadrant of the ellipse with the space curve $$\mathbf{r}(t)=a\cos(t)\hat{\mathbf{i}}+b\sin(t)\hat{\mathbf{j}} ~ \text{for } t\in[0,\pi/2].$$ The perimeter is $4(x+y)=4(a\cos(t)+b\sin(t))$. So, $$\frac{\mathrm{d}P}{\mathrm{d}t}=4(b\cos(t)-a\sin(t))$$ For a critical point we need $$b\cos(t)=a\sin(t)$$ Or, $$\frac{b}{a}=\tan(t)\implies t=\arctan\left(\frac{b}{a}\right)$$ So, $$x=a\cos\left(\arctan\left(\frac{b}{a}\right)\right)~;~y=b\sin\left(\arctan\left(\frac{b}{a}\right)\right)$$ With a little triangle diagram it's easy to show that the above is equivalent to what was found via Lagrange multipliers.

EDIT: See HERE for an improved visualization.

Answer 2

There must be a mistake in your derivation $\frac{dP}{dx}$. We have (in the positive quadrant): $$ y=b\sqrt{1-\frac{x^2}{a^2}}\text{ and so }P(x)=4(x+y)=4\left(x+b\sqrt{1-\frac{x^2}{a^2}} \right) $$ For the derivative, we get $$ \frac{dP(x)}{dx}=4-\frac{4bx}{a^2}\sqrt{1-\frac{x^2}{a^2}}^{-1} $$ and setting this equal to $0$, it follows that $$ 4=\frac{4bx}{a^2}\sqrt{1-\frac{x^2}{a^2}}^{-1} \Leftrightarrow 4\sqrt{1-\frac{x^2}{a^2}}=\frac{4bx}{a^2} \Leftrightarrow a\sqrt{a^2-x^2}=bx\\ \Rightarrow a^2(a^2-x^2)=b^2x^2\Leftrightarrow a^4=(a^2+b^2)x^2\\ \Rightarrow x=\frac{a^2}{\sqrt{a^2+b^2}} $$ The corresponding $y$ is $$ y=b\sqrt{1-\frac{a^2}{a^2+b^2}}=b\sqrt{\frac{b^2}{a^2+b^2}}=\frac{b^2}{\sqrt{a^2+b^2}} $$ and the corresponding perimeter is $$ P_\text{max}=4\left(\frac{a^2}{\sqrt{a^2+b^2}}+\frac{b^2}{\sqrt{a^2+b^2}}\right)=4\sqrt{a^2+b^2} $$