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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.

7 Answers 7

1

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.

  • 0
    In the fourth line, it should `Suppose ABCD is the rectangle.' instead of `Suppose ABCD is the ellipse.'2016-07-26
17

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 $A = 2ab$

  • 0
    I am sure that these comments should be added to the answer.2017-11-14
15

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

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

  • 0
    I have posted an answer here. http://math.stackexchange.com/questions/1793985/how-to-show-that-any-rectangle-in-ellipse-must-be-oriented-parallel-to-axes?rq=12016-07-26
3

${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 $4xy=2ab$.

  • 1
    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: http://math.stackexchange.com/questions/241741/simple-algebra-question-proving-a2b2-geqslant-2ab http://math.stackexchange.com/questions/320244/show-that-2-xy-x2-y2-for-x-is-not-equal-to-y http://math.stackexchange.com/questions/470221/prove-the-inequality-xy-leq-frac12x2y2 http://math.stackexchange.com/questions/943994/show-that-for-all-real-numbers-a-and-b-ab-le-1-2a2b22016-03-06
3

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

2

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 $2a^2\cdot b/a = 2ab$.