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Find the area of the largest rectangle that can be inscribed in the ellipse $$ \frac {x^2}{2} + \frac {y^2}{6} = 1$$

I know the answer is $x=1$ yields $Area = 4\sqrt{3}$.

I'm able to get $x=1$ by taking the derivative and setting equal to zero, but I don't understand where the $4$ in $4\sqrt{3}$ comes from?

brier
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    It is indeed the case that $2ab$ is the largest possible area of a quadrilateral inscribing an ellipse with semi-major axis $a$ and semi-minor axis $b$. Every quadrilateral of maximum area must be a parallelogram (and exactly one such quadrilateral is a rectangle). $$\phantom{a}$$ Anyway, you seem to be assuming that any rectangle of maximum area must have sides parallel to the axes of symmetry of the ellipse. I don't think this is trivial, although intuitively, this should be the case. – Batominovski Feb 07 '20 at 22:54
  • @WETutorialSchool You can’t inscribe a rectangle in an ellipse that isn’t parallel to the axes of symmetry of said ellipse. Even if that wasn’t the case, by OP’s posting history it seems clear to me that they’re in a beginning calculus course, certainly not the type where you would need to do any proofs. – Snacc Feb 07 '20 at 23:09
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    @connorlane Why can't one inscribe a rectangle in an ellipse with sides not parallel to the axes of symmetry? The OP's/your assertion hasn't been proven in any answer. I have a proof, but it involves geometric transformations and multi-variable calculus, which is probably not what the OP wants. I don't see an easy way to prove this, e.g., using elementary calculus or elementary geometry. – Batominovski Feb 07 '20 at 23:10
  • I’d expect OP’s teacher would have told them this fact (or, more likely, just showed them that this is how you optimize for area in an ellipse) . Again, it’s obvious he’s not supposed to prove that his answer is right. Proving the fact that you don’t need to consider those impossible rectangles isn’t a step to solving one of these problems. It’s likely a high school level calculus problem. It would be like if you asked algebra students to prove the fundamental theorem of algebra every time they found all the zeroes of a polynomial. – Snacc Feb 07 '20 at 23:39
  • https://math.stackexchange.com/questions/240192/find-the-area-of-largest-rectangle-that-can-be-inscribed-in-an-ellipse – lab bhattacharjee Feb 08 '20 at 01:48

2 Answers2

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Since your $x=1$ is correct, so I will continue from there. First, solve for y.

$$\frac{1^2}{2}+\frac{y^2}{6}=1$$ $$\frac{y^2}{6}=\frac{1}{2} $$ $$y^2=3$$ $$y=\sqrt{3}$$

Then, we double both of those values to find the side lengths. This is because if we didn’t do that, we would be calculating the area of the rectangle spanning from $(0,0)$ to $(1,\sqrt{3})$. By doubling them we get the area from $(-1,-\sqrt{3})$ to $(1,\sqrt{3})$

You get one side having length $2$, and the other with length $2\sqrt{3}$. Multiplying those gives the final answer of $4\sqrt{3}$

Snacc
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Your $x=1 $ gives you $y=\pm \sqrt 3$

The total area is four times the area in the first quadrant. Thus $$A=4xy=4(1)(\sqrt 3)$$

Mohammad Riazi-Kermani
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