# Thread: Area of a rectangle within an ellipse

1. ## Area of a rectangle within an ellipse

Find the dimensions of the largest rectangle wish dies parallel to the axes that can be inscribed in the ellipse x^2 + 4 y^2 = 4.

Okay, here's my work so far:
A=4xy
Rearrange for x so A=4y(sqrt(4-4y^2))
Use product rule, obtain derivative and set A' = 0
It eventually comes to A' = -32y^2/2 (sqrt (4-4y^2)) + 4 (sqrt (4-4y^2))
And I solve to get y = sqrt2/2, which is incorrect.
The answer for the dimensions is 2sqrt2 by sqrt2. I know I have to sub to get x back after, but is there some algebraic process which is incorrect here? Any help would be super.

2. Hello, cinematic!

Your work is correct . . . up to the punchline.
. . Did you forget to "double"?

Find the dimensions of the largest rectangle with sides parallel to the axes
that can be inscribed in the ellipse: $x^2 + 4 y^2 \:=\: 4$
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The upper half of the ellipse is: . $y \:=\:\sqrt{1-\frac{x^2}{4}}$

The area of the rectangle is: . $A \;=\;4xy$

We have: . $A \;=\;4x\left(1 - \tfrac{1}{4}x^2\right)^{\frac{1}{2}}$

Then: . $A' \;=\;4x\cdot\tfrac{1}{2}\left(1 - \tfrac{1}{4}x^2\right)^{\text{-}\frac{1}{2}}\! \left(\text{-}\tfrac{1}{2}x\right) + 4\left(1-\tfrac{1}{4}x^2\right)^{\frac{1}{2}}$

. . . $-\frac{x^2}{\sqrt{1-\frac{x^2}{4}}} + 4\sqrt{1 - \tfrac{x^2}{4}} \;=\;0 \quad\Rightarrow\quad \frac{x^2}{\sqrt{1-\frac{x^2}{4}}} \;=\;4\sqrt{1-\tfrac{x^2}{4}}$

. . . $x^2 \:=\:4\left(1 - \tfrac{x^2}{4}\right) \quad\Rightarrow\quad x^2 \:=\:4-x^2 \quad\Rightarrow\quad 2x^2 \:=\:4 \quad\Rightarrow\quad x^2 \:=\:2$

Hence: . $x \:=\:\sqrt{2} \quad\Rightarrow\quad y \:=\:\frac{\sqrt{2}}{2}$

Therefore: . $\begin{Bmatrix}\text{width} &=& 2x &=& 2\sqrt{2} \\ \text{height} &=& 2y &=& \sqrt{2} \end{Bmatrix}$

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### find the dimensions of the largest rectangle that can be inscribed in the ellipse

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