6. Let $x \geqslant 0, y \geqslant 0, 2x + y = 6$. Then the maximum value of $u = 4x^2 + 3xy + y^2 - 6x - 3y$ is ( ).
(A) $\frac{27}{2}$
(B) 18
(C) 20
(D) does not exist

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6. (B).

From the given, we have $y=6-2 x$. Substituting into $u=4 x^{2}+3 x y+y^{2}-$ $6 x-3 y$, we get
$$
u=2 x^{2}-6 x+18 \text {. }
$$

And $x \geqslant 0, y=6-2 x \geqslant 0$, so $0 \leqslant x \leqslant 3$.
When $x=0$ or $x=3$, $u_{\text {min }}=18$.