15. Let $x, y \in \mathbf{R}$ and satisfy $x^{2}+4 y^{2}=4$. Then the maximum and minimum values of $x^{2}+2 x y$ $+4 y^{2}$ are $\qquad$.
15.6,2.

Solution 1: Let $x=2 \cos \theta, y=\sin \theta, 0 \leqslant \theta<2 \pi$.
By $x^{2}+4 y^{2}=4$, then $x^{2}+2 x y+4 y^{2}=4+4 \sin \theta \cdot \cos \theta=4+2 \sin 2 \theta$. When $\sin 2 \theta=1$, $x^{2}+2 x y+4 y^{2}$ reaches its maximum value 6; when $\sin 2 \theta=-1$, $x^{2}+2 x y+4 y^{2}$ reaches its minimum value 2.
Solution 2: $\because 2|x y|=|x| \cdot|2 y| \leqslant \frac{x^{2}+4 y^{2}}{2}=2$, and when $|x|=|2 y|=\sqrt{2}$, the equality holds,
$\therefore-2 \leqslant 2 x y \leqslant 2$, i.e., $2 x y$ has a maximum value of 2 and a minimum value of -2. Therefore, $x^{2}+2 x y+4 y^{2}$ has a maximum value of 6 and a minimum value of 2.