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. Solution $1 \because x^{2}+4 y^{2}=4, \therefore$ let $x=2 \cos \theta, y=\sin \theta, 0 \leqslant \theta<2 \pi$, then we have $x^{2}+2 x y+4 y^{2}=4+4 \sin \theta \cdot \cos \theta=4+2 \sin 2 \theta$.
$\therefore$ 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 $|x|=|2 y|=\sqrt{2}$, equality holds;
$$
\therefore \quad-2 \leqslant 2 x y \leqslant 2 \text {. }
$$

That is, $2 x y$ has a maximum value of 2 and a minimum value of -2.
Therefore, the maximum value of $x^{2}+2 x y+4 y^{2}$ is 6, and the minimum value is 2.