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: $\displaystyle x^2 + 4 y^2 \:=\: 4$ Code:

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The upper half of the ellipse is: .$\displaystyle y \:=\:\sqrt{1\frac{x^2}{4}}$
The area of the rectangle is: .$\displaystyle A \;=\;4xy$
We have: .$\displaystyle A \;=\;4x\left(1  \tfrac{1}{4}x^2\right)^{\frac{1}{2}}$
Then: .$\displaystyle 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}} $
. . . $\displaystyle \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}} $
. . . $\displaystyle x^2 \:=\:4\left(1  \tfrac{x^2}{4}\right) \quad\Rightarrow\quad x^2 \:=\:4x^2 \quad\Rightarrow\quad 2x^2 \:=\:4 \quad\Rightarrow\quad x^2 \:=\:2$
Hence: .$\displaystyle x \:=\:\sqrt{2} \quad\Rightarrow\quad y \:=\:\frac{\sqrt{2}}{2}$
Therefore: .$\displaystyle \begin{Bmatrix}\text{width} &=& 2x &=& 2\sqrt{2} \\ \text{height} &=& 2y &=& \sqrt{2} \end{Bmatrix}$