6・114 Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=20$, find the maximum and minimum values of $x^{2}-x y+y^{2}$.

Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=20$, find the maximum and minimum values of $x^{2}-x y+y^{2}$.
[Solution] Squaring both sides of $|5 x+y|+|5 x-y|=20$, we get
$$
25 x^{2}+y^{2}+\left|25 x^{2}-y^{2}\right|=200 \text {. }
$$

If $|5 x| \geqslant|y|$, then we have
$$
50 x^{2}=200, x= \pm 2,|y| \leqslant 10 .
$$

If $|5 x| \leqslant|y|$, then we have
$$
2 y^{2}=200, y= \pm 10,|5 x| \leqslant 10,|x| \leqslant 2 .
$$

Therefore, the graph of the equation is a rectangle $A B C D$, where
$$
A(2,-10), B(2,10), C(-2,10), D(-2,-10) \text {. }
$$

By symmetry, we only need to consider on the sides $A B$ and $B C$, and it is easy to see that
$$
\text { On } \begin{aligned}
A B \text {, } Q & =x^{2}-x y+y^{2}=4-2 y+y^{2}, \\
3 & \leqslant Q \leqslant 124 ;
\end{aligned}
$$

On $B C$, $84 \leqslant Q \leqslant 124$.
$$
Q_{\text {maximum }}=124, Q_{\text {minimum }}=3 \text {. }
$$