Variation 2 (Excerpt from the 2008 National Exam II, Science Question 21) Let the center of the ellipse be at the origin of the coordinate system, with $A(2,0), B(0,1)$ being two of its vertices. The line $y=k x(k>0)$ intersects $A B$ at point $D$, and intersects the ellipse at points $E$ and $F$. Find the maximum value of the area of quadrilateral $A E B F$.
For the quadrilateral $A O B F$ doubled, let point $F(x, y)(x>0)$, so the area of quadrilateral $A E B F$ is $S=2\left(S_{\triangle O B F}+S_{\triangle O A F}\right)=x+2 y$. Also, since $1=\frac{x^{2}}{4}+y^{2}=\frac{x^{2}}{4}+\frac{(2 y)^{2}}{4} \geqslant \frac{(x+2 y)^{2}}{4+4}=\frac{(x+2 y)^{2}}{8}$, thus $x+2 y \leqslant 2 \sqrt{2}$. Therefore, the maximum area of quadrilateral $A E B F$ is $2 \sqrt{2}$, which is achieved when and only when $x / 4=2 y / 4$, i.e., $x=\sqrt{2}$.

Solution: According to the problem, the equation of the ellipse is $\frac{x^{2}}{4}+y^{2}=1,|B O|=1$, $|A O|=2$. By the symmetry of the ellipse, it is easy to know that the area of quadrilateral $A E B F$ is

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