1. As shown in Figure 3, in the Cartesian coordinate system $x O y$, the diagonals of rhombus $O A B C$ intersect at point $M\left(x_{0}, y_{0}\right)$, and the hyperbola $y=\frac{k}{x}(x>0)$ passes through points $C$ and $M$. If $2 \leqslant x_{0} \leqslant 4$, then the maximum value of the area of rhombus $O A B C$ is $\qquad$
II, $1.24 \sqrt{2}$.
From the problem, we have $y_{0}=\frac{k}{x_{0}}$.
Let's assume $C\left(\frac{x_{0}}{2}, \frac{2 k}{x_{0}}\right), A\left(\frac{3 x_{0}}{2}, 0\right)$.
Since $O A=O C$, we have,
$\left(\frac{3 x_{0}}{2}\right)^{2}=\left(\frac{x_{0}}{2}\right)^{2}+\left(\frac{2 k}{x_{0}}\right)^{2}$
$\Rightarrow k=\frac{\sqrt{2}}{2} x_{0}^{2}$.
Therefore, the area of the rhombus $O A B C$ is
$$
S=x_{A} y_{C}=\frac{3 x_{0}}{2} \cdot \frac{2 k}{x_{0}}=3 k=\frac{3 \sqrt{2}}{2} x_{0}^{2} \text {. }
$$

When $x_{0}=4$, $S$ reaches its maximum value $24 \sqrt{2}$.