6. Let point $A(2,0)$, and $B$ be a point on the elliptical arc
$$
\frac{x^{2}}{4}+\frac{y^{2}}{3}=1(x>0, y>0)
$$

Draw a perpendicular from point $B$ to the $y$-axis, and let the foot of the perpendicular be $C$. Then the maximum value of the area of quadrilateral $O A B C$ is $\qquad$
6. $\frac{9}{4}$.

Let $B(2 \cos \theta, \sqrt{3} \sin \theta)\left(\theta \in\left(0, \frac{\pi}{2}\right)\right)$.
Then, $C(0, \sqrt{3} \sin \theta)$.
Thus, $S_{\text {quadrilateral } A B C O}$
$$
\begin{array}{l}
=2 \cos \theta \cdot \sqrt{3} \sin \theta+\frac{1}{2}(2-2 \cos \theta) \sqrt{3} \sin \theta \\
=\frac{\sqrt{3}}{2} \sin 2 \theta+\sqrt{3} \sin \theta .
\end{array}
$$

To find the maximum and minimum values of the area of quadrilateral $A B C O$, we need to differentiate the above expression with respect to $\theta$ and set the result to 0, solving for the critical points, i.e.,
$$
\begin{array}{l}
2 \cos ^{2} \theta+\cos \theta-1=0 \\
\Rightarrow \theta=\frac{\pi}{3} \text { or } 0 \text { (discard). }
\end{array}
$$

Upon verification, when $\theta=\frac{\pi}{3}$, $S_{\text {quadrilateral } A B C O}$ reaches its maximum value $\frac{9}{4}$.