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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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up vote 6 down vote accepted

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

$${1=\frac{{ x }^{ 2 }}{{ a }^{ 2 }} + \frac {{ y }^{ 2 }} {{ b }^{ 2 }}} \ge \frac{2 { xy }}{{ ab }} $$

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

i.e. :max of $xy =ab/2$, so $4xy=2ab$.

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The first inequality used in this post is $u^2+v^2\ge 2uv$ for $u=x/a$ and $v=y/b$. See, for example:………… – Martin Sleziak Mar 6 at 10:03

Stretching the plane in a given direction is an operation that preserves ratios of areas. So:

  • Stretch the plane by a factor of $a/b$ in the $y$-direction, to transform the ellipse to a circle with radius $a$.

  • Inscribe the largest possible rectangle inside this circle, which turns out to be a square of area $2a^2$. Align this square with the $xy-$axes.

  • Stretch the plane by a factor of $b/a$, to return the ellipse to its original size and shape. The resulting rectangle has area $2a^2\cdot b/a = 2ab$.

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Parametric form of ellipse

$$ x = a \cos t,\; y = b \sin t, \; A = 4 x y = 2 a b \sin (2 t);$$

Maximum area occurs for rectangle cutting by radial straight lines at $ t= \pm 45^0 $ through origin, a $\dfrac2{\pi}$ fraction of the ellipse area. Due to two axis symmetry slanted orientations of rectangles can be ruled out.

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