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So say I have an ellipse defined like this:

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

I have to find the largest possible area of an inscribed rectangle.

So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefine $y$ in terms of $x$:

$$\frac{y^2}{4} = 1 - \frac{x^2}{9}$$ $$y^2 = 4 - \frac{4x^2}{9}$$ $$y = \sqrt{4 - \frac{4x^2}{9}}$$

So the area function is now:

$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}$$ $$A' = \frac{4x}{2 \cdot \frac{-8x}{9}} + \sqrt{4 - \frac{4x^2}{9}} \cdot 4$$

So this track seems too difficult, so I'd like to find another approach. Can I square the area first, find the derivative of that to solve for $x$?

So the $\text{Area} = 4x \cdot \sqrt{4 - \frac{4x^2}{9}}$

Is this valid?

$$\text{Area}^2 = 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right)$$

$$= 64x^2 - \frac{64x^4}{9}$$

Derivative:

$$ \frac{d}{dx} \text{Area}^2 = 128x - \frac{256x^3}{9}$$ $$128x\left(1-\frac{2x^2}{9}\right)$$

So critical values: $x = 0, \frac{3}{\sqrt{2}}$

because the derivative equals $0$ when:

$$2x^2 = 9$$ $$x = \frac{3}{\sqrt{2}}$$

Plugging this value of $x$ into $y$ we get that $y = \sqrt{2}$ so the $\text{Area}$ is $3$.

Is this valid? If so why? Does squaring not cause any problems?

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    $\begingroup$ If you have a set of positive numbers the largest one has the largest square. So in your case squaring causes no problem. $\endgroup$ – Ethan Bolker Mar 10 '18 at 16:10
  • $\begingroup$ By your method area should be $4xy$ $\endgroup$ – sku Mar 10 '18 at 16:22
  • $\begingroup$ Your A' shown above is not correct. $\endgroup$ – Mathew Mahindaratne Mar 10 '18 at 16:28
  • $\begingroup$ Even after correcting that missing factor of 4, you only proved that is the rectangle with the largest area among those with sides parallel to the axes of the ellipse. It may seem obvious that others have a still smaller area, but showing that rigorously might be a bit of a challenge. $\endgroup$ – Professor Vector Mar 10 '18 at 16:33
  • $\begingroup$ Area, $A=4xy$ so it should be $12$. $\endgroup$ – Mathew Mahindaratne Mar 10 '18 at 17:14
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Yes, it is valid. You want to determine the maximum of a non-negative function $f$. But, since $f$ is non-negative, asserting the $\max f=M$ is equivalent to asserting that $\max f^2=M^2$. Besides, $f(x)=M\iff f^2(x)=M^2$. So, the points at which the functions $f$ and $f^2$ attain their maximal value are the same.

Note that $f$ being non-negative is essential. If $f(x)=x$, with $x\in[-2,1]$, then $f$ has a maximum at $1$, whereas $f^2$ hasn't (its maximum value is attained at $-2$).

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  • $\begingroup$ Is my math correct? $\endgroup$ – Jwan622 Mar 10 '18 at 16:50
  • $\begingroup$ @Jwan622 Almost. You got $x=\frac3{\sqrt2}$, which is correct. Therefore $y=\sqrt2$ (here you are correct again). But the conclusion is that the area is $12$, not $3$. $\endgroup$ – José Carlos Santos Mar 10 '18 at 16:54
  • $\begingroup$ don't you just multiply x by y? $\endgroup$ – Jwan622 Mar 10 '18 at 22:34
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We all know that the quadrangle of maximal area inscribed in a disc is a square, and that this square covers ${2\over\pi}$ of the area of the disc. Your ellipse has semiaxes $2$ and $3$, hence area $6\pi$. The largest rectangle inscribed in this ellipse therefore has area $$\leq{2\over\pi}\cdot 6\pi=12\ ,$$ and this value is attained for a suitable axis-aligned rectangle.

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Starting from $$ \left\{ \matrix{ {{x^{\,2} } \over 9} + {{y^{\,2} } \over 4} = 1 \hfill \cr A = 4xy \hfill \cr} \right. $$ which gives the intercepts between the ellipse and a hyperbola, you could have simply substitute $y=A/(4x)$, to get $$ {{x^{\,2} } \over 9} + {{A^{\,2} } \over {64x^{\,2} }} = 1 $$ i.e. $$ 64x^{\,4} - 9 \cdot 64x^{\,2} + 9A^{\,2} = 0 $$ put the discriminant to be null, and get $A$ $$ \Delta = \left( {9 \cdot 64} \right)^{\,2} - 4 \cdot 64 \cdot 9A^{\,2} = 0\quad \Rightarrow \quad A^{\,2} = 9 \cdot 16\quad \Rightarrow \quad A = 12 $$ and since the discriminant is null, from the above it follows $$ x^{\,2} = {{9 \cdot 64} \over {2 \cdot 64}}\quad \Rightarrow \quad x = {3 \over 2}\sqrt 2 \quad \Rightarrow \quad y = A/4x = \sqrt 2 $$

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I agree with José Carlos Santos, yet I don't agree with your $A^2$ derivative. So, I like to show my way of solving with $A^2$. $$Area =A= 4x \cdot \left(4 - \frac{4x^2}{9}\right)^{\frac12}$$ $$A^2= 16x^2 \cdot \left(4 - \frac{4x^2}{9}\right) = 64x^2 - \frac{64x^4}{9}$$

Thus, derivative of $A^2$: $$ \frac1A\cdot\frac{dA}{dx} = 128x - \frac{256x^3}{9}=128x\left(1-\frac{2x^2}{9}\right)$$ Thus, $$ \frac{dA}{dx} =A\cdot 128x\cdot \left(1-\frac{2x^2}{9}\right)$$

Since $A\ne 0$, the critical values are: $x = 0$ or $x=\pm \frac{3}{\sqrt{2}}$ because $\frac{dA}{dx}$ equals $0$ at critical points. Note that, as Professor Vector pointed out, "the rectangle with the largest area among those with sides parallel to the axes of the ellipse." There are two sides parallel to the $y$-axis, and the points of $x=+\frac{3}{\sqrt{2}}$ and $x=-\frac{3}{\sqrt{2}}$ are where these parallel sides meet the $x$-axis. Consequently, the maximum area of rectangle is: $$A_{max}=$4\left(\frac{3}{\sqrt{2}}\right) \cdot \left(4 - \frac{4\left(\frac{3}{\sqrt{2}}\right)^2}{9}\right)^{\frac12}=12$$

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$$ \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \quad \Rightarrow \quad \dfrac{2x}{9}dx + \dfrac{y}{2}dy = 0 \quad (1) $$ $$ A = 4xy \quad \Rightarrow \quad 0 = dA = 4ydx + 4xdy \quad (2) $$ Therefore, from (1) and (2) $$ \dfrac{dy}{dx} = -\dfrac{4x}{9y} \quad \text{and} \quad \dfrac{dy}{dx} = -\dfrac{y}{x} \quad \Rightarrow \quad 4x^2 = 36\dfrac{y^2}{4} = 36\biggl(1 - \dfrac{x^2}{9}\biggr) \quad \Rightarrow \quad x = \pm \dfrac{3}{\sqrt{2}} $$ The maximum occurs for $x = 3/\sqrt{2}$.

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