6. If non-negative real numbers $x, y$ satisfy
$$
x^{2}+4 y^{2}+4 x y+4 x^{2} y^{2}=32 \text {, }
$$

then the maximum value of $\sqrt{7}(x+2 y)+2 x y$ is $\qquad$
6. 16 .

Let $\sqrt{7}(x+2 y)+2 x y=t(t \geqslant 0)$. Then
$$
\begin{array}{l}
\sqrt{7}(x+2 y)=t-2 x y \geqslant 0 . \\
\Rightarrow 7(x+2 y)^{2}=(t-2 x y)^{2} . \\
\text { By } x^{2}+4 y^{2}+4 x y+4 x^{2} y^{2}=32 \\
\Rightarrow x^{2}+4 y^{2}+4 x y=(x+2 y)^{2} \\
\quad=32-4 x^{2} y^{2} .
\end{array}
$$
(1) - (2) gives
$$
\begin{array}{l}
7\left(32-4 x^{2} y^{2}\right)=(t-2 x y)^{2} \\
\Rightarrow 32 x^{2} y^{2}-4 t x y-32 \times 7+t^{2}=0 .
\end{array}
$$

Since the above equation in terms of $x y$ must have a solution, we have
$$
\begin{array}{l}
\Delta=16 t^{2}-4 \times 32\left(t^{2}-32 \times 7\right) \geqslant 0 \\
\Rightarrow t^{2} \leqslant 16 \times 16 .
\end{array}
$$

Since $t \geqslant 0$, therefore, $0 \leqslant t \leqslant 16$.
Thus, the maximum value of $t$ is 16, which is the required maximum value.