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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:

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

Is this the right track? Was there something simpler I could have done? this looks gnarly? Can someone help me finish this up?

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

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

Is this valid?

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

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

Derivative:

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

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 Area is 3.

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

Jwan622
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  • Your first term in $A'$ is incorrect – A. Goodier Mar 09 '18 at 21:25
  • @rbird I don't see why.... – Jwan622 Mar 09 '18 at 22:28
  • If $y$ is a function of $x$, derivative of $y^2= 2 y y'^{'} $. Derivatives of $y,y^2$ vanish together. Avoid cumbersome calculation by squaring. Also applying Chain Rule to the squared quantity, it goes much faster $ \dfrac{x^2}{1-x^2/9}=\dfrac{-2x}{-2x/9}=9 \implies x=\pm \dfrac{3}{\sqrt2}$ – Narasimham Mar 10 '18 at 06:06

3 Answers3

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You are over-complicating it. Affine maps preserve the ratios of areas and, in a circle, the inscribed squares are pretty obviously the largest inscribed quadrilaterals. Their area is $\frac{2}{\pi}$ times the area of the circle. By applying $\varphi:(x,y)\mapsto(ax,by)$ the ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is mapped into a circle. By applying $\varphi^{-1}$ we get that

A. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the largest inscribed rectangle (which is also one of the largest inscribed quadrilaterals) is symmetric with respect to the axis of the ellipse and its area equals $2ab$.

By the very same principle,

B. In the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the area of the largest inscribed triangle equals $\frac{3\sqrt{3}}{4}ab$.

C. In a triangle $ABC$ the largest inscribed ellipse is tangent to the sides at their midpoints and the center of such ellipse lies at the centroid of $ABC$. This is the Steiner inellipse of $ABC$.

Community
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Jack D'Aurizio
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  • I don't know these principles yet and I have to brute force it. – Jwan622 Mar 09 '18 at 21:43
  • @Jwan622: I believe that affine maps preserve the ratios of areas is a much more elementary concept than $\frac{d}{dx}$. Now that you are aware of it, why not to use it? – Jack D'Aurizio Mar 09 '18 at 21:45
  • How can we convince ourselves that the largest rectangle has sides parallel to the main ellipse axis ? –  Mar 09 '18 at 22:23
  • @YvesDaoust: by applying $\varphi^{-1}$ to all the inscribed squares in the circle. – Jack D'Aurizio Mar 09 '18 at 23:05
  • @JackD'Aurizio: if the rectangle and the ellipse are not aligned, no affine map can lead to a square inscribed in a circle. But such a configuration cannot be rejected a priori. –  Mar 10 '18 at 00:28
  • Maybe I did not explain me clearly. We know all the quadrilaterals with maximal area inscribed in a circle. By pulling them back, we get all the quadrilaterals with maximal area inscribed in a ellipse. One of them is a rectangle. In particular it is the largest inscribed rectangle. – Jack D'Aurizio Mar 10 '18 at 01:38
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Hint...it would be a lot easier if you wrote $$(x,y)=(3\cos\theta,2\sin\theta)$$ then obtain an expression for the area of the rectangle in terms of $\theta$ and differentiate...

David Quinn
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I think you are in right track but your expression for the first derivative of $A$ need a little touch. I put it in more clear way:

$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}=4x \left(4-\frac{4x^2}{9}\right)^{1/2}$$ Thus, $$A' = 4 \left(4-\frac{4x^2}{9}\right)^{1/2} + 4x \left(4-\frac{4x^2}{9}\right)^{-1/2}\left(\frac{-8x}{9} \right)=4 \left(4-\frac{4x^2}{9}\right)^{1/2} - \frac{32x^2}{9} \left(4-\frac{4x^2}{9}\right)^{-1/2}=\frac{36 \left(4-\frac{4x^2}{9}\right) - 32x^2} {9\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{4 \left(36-4x^2\right) - 288x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}=\frac{144 - 304x^2} {81\left(4-\frac{4x^2}{9}\right)^{1/2}}$$ To find the maximum for $A$, set $A'=0$ since $-3\le x\le +3$, and hence $\left(4-\frac{4x^2}{9}\right)^{1/2}\ne0$. Therefore, $144 - 304x^2=0$ when $A'-0$ and $x=\pm\sqrt\frac{19}{9}$. $$A_{max}=4\left(\sqrt\frac{19}{9}\right) \left( \sqrt{4 - \frac{4\left(\sqrt\frac{19}{9}\right)^2}{9}}\right)=8\left(\sqrt\frac{19}{9}\right) \left( \frac{\sqrt{62}}{9}\right)=\frac{8\sqrt{1178}}{27}$$

Note: The $\pm x$ values for rectangle are symmetrical points on major axis of ellipse where vertical legs of rectangle meet.

Mathew Mahindaratne
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