Area of largest inscribed rectangle in an ellipse. Can I take the square of the area to simplify calculations?

Question

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?

Answer 1

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 $-2$).

Answer 2

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: $$A_{max}=$4\left(\frac{3}{\sqrt{2}}\right) \cdot \left(4 - \frac{4\left(\frac{3}{\sqrt{2}}\right)^2}{9}\right)^{\frac12}=12$$

Answer 3

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.

Answer 4

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 $$ x^{\,2} = {{9 \cdot 64} \over {2 \cdot 64}}\quad \Rightarrow \quad x = {3 \over 2}\sqrt 2 \quad \Rightarrow \quad y = A/4x = \sqrt 2 $$

Answer 5

$$ \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \quad \Rightarrow \quad \dfrac{2x}{9}dx + \dfrac{y}{2}dy = 0 \quad (1) $$ $$ A = 4xy \quad \Rightarrow \quad 0 = dA = 4ydx + 4xdy \quad (2) $$ 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 $x = 3/\sqrt{2}$.