# Problem 5.

Find the smallest and largest value of the expression $|x+2|+|y+3|$ under the condition $(|x|-3)^{2}+(|y|-2)^{2}=1$.
Solution. The graph of the equation $(|x|-3)^{2}+(|y|-2)^{2}=1$ is the union of four unit-radius circles centered at points $(\pm 3, 2)$, $(\pm 3, -2)$. The graph of the equation $|x+2|+|y+3|=a$ (for $a>0$) is a family of squares centered at the point $(-2, -3)$, with diagonals parallel to the coordinate axes and equal to $2a$.

The smallest value of $a$ corresponds to the external tangency of the side of the square to the nearest circle, located in the third quadrant. The largest value of $a$ corresponds to the internal tangency of the side of the square to the farthest circle, located in the first quadrant. We obtain:

$\min a=(\sqrt{2}-1) \cdot \sqrt{2}=2-\sqrt{2}$,

$\max a=(5 \sqrt{2}+1) \cdot \sqrt{2}=10+\sqrt{2}$.

Answer: $2-\sqrt{2} ; 10+\sqrt{2}$.