6. Given in the Cartesian coordinate system $x O y$, the line $y=2 k x+3-4 k$ intersects the positive semi-axis of $x$ and the positive semi-axis of $y$ at points $A$ and $B$ respectively, and $P$ is a point on the line segment $A B$. $P M \perp x$ axis at point $M$, and $P N \perp y$ axis at point $N$. Then the maximum value of the area of rectangle $O M P N$ is at least ( ).
(A) 3
(B) 4
(C) 6
(D) 8
6.C.

Let the coordinates of point $P$ be $\left(x_{0}, y_{0}\right)$, and the area of rectangle OMPN be $S$. Then $x_{0}>0, y_{0}>0, S=x_{0} y_{0}$.
Since point $P\left(x_{0}, y_{0}\right)$ lies on $y=2 k x+3-4 k$, we have $y_{0}=2 k x_{0}+3-4 k$.
Thus, $S=x_{0}\left(2 k x_{0}+3-4 k\right)=2 k x_{0}^{2}+(3-4 k) x_{0}$.
Therefore, $S_{\text {max }}=\frac{4 \times 2 k \times 0-(3-4 k)^{2}}{4 \times 2 k}$, which means $16 k^{2}-\left(24-8 S_{\text {max }}\right) k+9=0$.
Since $k$ is a real number, we have
$$
\Delta=\left[-\left(24-8 S_{\text {max }}\right)\right]^{2}-4 \times 16 \times 9 \geqslant 0 \text {. }
$$

Solving this, we get $S_{\text {max }} \geqslant 6$ or $S_{\text {max }} \leqslant 0$ (discard).
When $S_{\text {max }}=6$, $k=-\frac{3}{4}$.