Ukrajina 2008

Expression $f=x^2-6x+y^2-6y=(x-3{)}^2+(y-3{)}^2-18=R^2-18$ (fig.6) is at maximum (minimum) if expression $R^2=(x-3{)}^2+(y-3{)}^2$ being an equation of the circle of radius $R$ with the center at point $(3,3)$ is at maximum (minimum). The graph of equation $|x+y|+|x-y|=1$ is a square formed by lines $x=\pm \frac{1}{2}$, $y=\pm \frac{1}{2}$, see fig.6.



Among all the circles intersecting the square, the circle passing through the point $B(-\frac{1}{2},-\frac{1}{2})$ has the maximal radius, and the circle passing through the point $A(\frac{1}{2},\frac{1}{2})$ has the minimal radius. Thus we find $R^2=(\frac{7}{2}{)}^2+(\frac{7}{2}{)}^2=\frac{49}{2}$ and $f_{max}=\frac{49}{2}-18=\frac{13}{2}$ for maximum, and $R^2=(\frac{5}{2}{)}^2+(\frac{5}{2}{)}^2=\frac{25}{2}$, $f_{min}=\frac{25}{2}-18=-\frac{11}{2}$ for minimum.