# Optimization of the Area of a rectangle with regards to 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.

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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

$$x=\frac{a}{\sqrt2}=\frac{a\sqrt2}2\quad\text{and}\quad y=\frac{b}{\sqrt2}=\frac{b\sqrt2}2\;.$$

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+1 Very nicely stated, Brian! –  amWhy Nov 18 '12 at 22:40
@amWhy: Thanks! –  Brian M. Scott Nov 18 '12 at 22:40
Thanks! This is more closely related to how I was thinking of it. –  Gabby Nov 18 '12 at 22:51
@Gabby: You’re welcome. I thought that it might be, from the way you wrote the question. –  Brian M. Scott Nov 18 '12 at 22:54

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 $$A = 2ab$$

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I don't get how you got (±acos(θ),±bsin(θ)). –  Gabby Nov 18 '12 at 22:32
@Gabby Any point on the ellipse if given by $(a \cos(\theta), b \sin(\theta))$. Any quadrilateral inscribed in an ellipse will have coordinates $(a \cos(\theta_1), b \sin(\theta_1))$, $(a \cos(\theta_2), b \sin(\theta_2))$, $(a \cos(\theta_3), b \sin(\theta_3))$ and $(a \cos(\theta_4), b \sin(\theta_4))$. The fact that it is a rectangle enforces that $\theta_2 = \pi - \theta_1$, $\theta_4 = - \theta_1$ and $\theta_3 = \pi + \theta_1$ –  user17762 Nov 18 '12 at 22:33
Thank you very much :) –  Gabby Nov 18 '12 at 22:50

$${1={ x }^{ 2 } /{ a }^{ 2 } + { y }^{ 2 } /{ b }^{ 2 }} >=2 { xy }/{ ab }$$

when and only when $${ x }/{ a } = { y }/{ b }$$ , the max is got

ie:max of xy =ab/2,so 4xy=2ab

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