绝对值方程(|x+1|+|x-2|)(|y+1|+|y-2|)(|z-3|+|z+1|)=36 求x+2y+3Z的最大、最小值

(|x+1|+|x-2|) >= (|(x+1)-(x-2)|)=3
(|y+1|+|y-2|) >= (|(y+1)-(y-2)|)=3
(|z-3|+|z+1|) >= (|(z-3)-(z+1)|)=4
在满足上述条件的情况下,36只能分解为3x3x4
则必有
(|x+1|+|x-2|)=3,当-1<=x<=2时,此条件满足
(|y+1|+|y-2|)=3,当-1<=y<=2时,此条件满足
(|z-3|+|z+1|)=4,当-1<=z<=3时,此条件满足
则
最小值：x=y=z=-1,x+2y+3z = -6
最大值：x=y=2,z=3,x+2y+3z = 15