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

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 $4\sqrt{3}$
