, 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.