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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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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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  • $\begingroup$ Thanks! This is more closely related to how I was thinking of it. $\endgroup$ – Gabby Nov 18 '12 at 22:51
  • $\begingroup$ @Gabby: You’re welcome. I thought that it might be, from the way you wrote the question. $\endgroup$ – Brian M. Scott Nov 18 '12 at 22:54
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    $\begingroup$ Why is it clear that the area is 4xy? It is not even clear that the vertices of the rectangle should be $(x,y)$, $(-x,y)$, $(-x,-y)$ and $(x,-y)$. This requires a proof. This is taken for granted in all the solutions. $\endgroup$ – user157986 Jul 24 '16 at 6:43
  • $\begingroup$ @user157986: It's taken for granted because unless $x=y$, that's the only way to inscribe a rectangle in the ellipse. This is tedious but straightforward to verify with a bit of analytic geometry, and also follows from the fact that the rectangle is symmetric about both coordinate axes. Moreover, the OP, rigorously or otherwise, had evidently come to that conclusion already, so I was dealing with the part that was causing difficulty. $\endgroup$ – Brian M. Scott Jul 24 '16 at 7:30
  • $\begingroup$ @Brian M. Scott: Sorry if I am a persistent cuss. This is precisely the point I am trying to understand. There are answers here: math.stackexchange.com/questions/1793985/… But they don't look convincing. Even if it is tedious, can you just explain the main steps? I will try to construct the proof myself. $\endgroup$ – user157986 Jul 25 '16 at 17:07
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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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    $\begingroup$ I don't get how you got (±acos(θ),±bsin(θ)). $\endgroup$ – Gabby Nov 18 '12 at 22:32
  • $\begingroup$ @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 $\theta_3 = \pi + \theta_1$ $\endgroup$ – user17762 Nov 18 '12 at 22:33
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    $\begingroup$ @user17762 What is $\theta$ here? $\endgroup$ – user157986 Jul 24 '16 at 6:39
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    $\begingroup$ eccentric angle $\endgroup$ – mehulmpt Mar 16 '17 at 8:50
  • $\begingroup$ I am sure that these comments should be added to the answer. $\endgroup$ – beroal Nov 14 '17 at 8:07
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$${1=\frac{{ x }^{ 2 }}{{ a }^{ 2 }} + \frac {{ y }^{ 2 }} {{ b }^{ 2 }}} \ge \frac{2 { xy }}{{ ab }} $$

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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let L and H be the length and breadth of the required rectangle respectively

$$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$

$$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$

$$H=\frac{b}{a}\sqrt{4a^2-L^2}$$

Area=L*H

$$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$

$$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0$$

$$\frac{b*(4a^2-2L^2)}{a\sqrt{4a^2-L^2}}$$

$$=> 4a^2-2L^2=0$$

$$=2a^2=L^2$$ $$L=a\sqrt{2}$$ $$H=b\sqrt{2}$$ $$Area=L*H=2ab$$

$$\frac{d^2A}{dL^2}=\frac{\sqrt{4a^2-L^2}*(-4L)-\frac{4a^2-2L^2}{2\sqrt{4a^2-L^2}}}{4a^2-L^2}$$

Putting L=$a\sqrt{2}$ $$\frac{d^2A}{dL^2}=\frac{-a\sqrt{2}(4a\sqrt{2})-\frac{0}{2\sqrt{4a^2-2a^2}}}{4a^2-2a^2}$$

$$=\frac{-8a^2}{2a^2}$$

-4<0.

Therefore the area is max

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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

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

EDIT1:

Maximum area occurs for rectangle cutting by radial straight lines at $ t= \pm 45^0 $ through origin. The ellipse area is $\dfrac{2}{\pi}$ fraction of the enveloping rectangle area . The ellipse passes through rectangle corners $ \frac{a}{\sqrt2},\frac{b}{\sqrt2}.$

Due to two axis symmetry slanted orientations of rectangles can be ruled out.

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  • $\begingroup$ I don't understand what you say in the last line. How does this follow from symmetry? $\endgroup$ – user157986 Jul 24 '16 at 6:45
  • $\begingroup$ @user157986: Inscribe the rectangle. Reflect ellipse and rectangle in one axis of symmetry of the rectangle, then in the other. The rectangle ends up in its original position, and since the net effect of the two reflections is to rotate the figure $180$ degrees, so does the ellipse. This is possible iff the axes of symmetry of the ellipse are the coordinate axes. $\endgroup$ – Brian M. Scott Jul 24 '16 at 7:39
  • $\begingroup$ @user 157986 Reflection symmetry as explained by Brian M. Scott $\endgroup$ – Narasimham Jul 24 '16 at 15:00
  • $\begingroup$ @Brian Scott: In the last line, I suppose you meant that the `This is possible iff the axes of symmetry of the rectangle are the coordinate axes.', I suppose my geometric visualisation is bad. It is not clear to me why the centre of the ellipse and the centre of the rectangle(point of intersection of the diagonals) are the same. If they are not the same, the ellipse will not end up in its original position. $\endgroup$ – user157986 Jul 25 '16 at 16:32
  • $\begingroup$ @Narasimham: The $t$ in the parametrisation is not the polar angle of the point on the ellipse at the centre, but what is known as eccentric anomaly. See en.wikipedia.org/wiki/Eccentric_anomaly. $\endgroup$ – user157986 Jul 25 '16 at 16:35
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Using the same technique shown here, we can orthogonally project the desired rectangle to the inscribed rectangle in the unit circle with maximal area (i.e. $2$, consider the inscribed square with sidelength $\sqrt{2}$).

Let the maximal area of our rectangle be $\mathcal{A}$. Then, by preservation of area ratios,

$$\begin{align*} \frac{\mathcal{A}}{\text{ Area of ellipse}} &=\frac{\text{Area of maximal square}}{\text{ Area of circle}} \\\frac{\mathcal{A}}{ab\pi} &= \frac{2}{\pi}\\ \implies \mathcal{A} &=2ab\end{align*}$$ which is our answer.

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