3. Let $P(x, y)$ be a point on $|5 x+y|+|5 x-y|=20$, find the maximum and minimum values of $x^{2}-x y+y^{2}$.
(1993 Macau Competition Problem)
3. Squaring both sides of $|5 x+y|+|5 x-y|=20$, and simplifying, we get $25 x^{2}+y^{2}+\left|25 x^{2}-y^{2}\right|=200$. If $|5 x|>|y|$, then we get $50 x^{2}=200, x= \pm 2,|y| \leqslant 10$; if $|5 x| \leqslant|y|$, then we get $2 y^{2}=200, y= \pm 10,|5 x| \leqslant 10,|x| \leqslant$ 2. Therefore, the graph of the given equation is a rectangle $A B C D$, where $A(2,-10), B(2,10), C(-2,10), D(-2,-10)$.

By symmetry, we only need to consider the sides $A B$ and $B C$, and it is easy to see that on $A B$, $Q=x^{2}-x y+y^{2}=4-2 y+y^{2}, 3 \leqslant Q \leqslant$ 124, and on $B C$, $84 \leqslant Q \leqslant 124$, so $Q_{\text {max }}=124, Q_{\text {min }}=3$.