若x，y满足|x-1|+2|y+1|≤2，则M=2x²+y²-2x的最小值为

更新时间2018-04-22 16:06:31

求解

|x-1|+2|y+1|≤2

|y+1|≤1-½|x-1|

y≤-½|x-1|      -1≤y≤0

y≥½|x-1|-2 y  -2≤y≤-1

∵y≤-½|x-1|≤0→y²≥¼|x-1|²

∴2x²+y²-2x≥2x²+¼|x-1|²-2x=2.25x²-2.5x+¼=2.25(x-5/9)²-4/9≥-4/9

即：当x=5/9 y=-2/9时，取得最小值-4/9

绝对值(x-1)<=2,即-1<=x<=3；绝对值2（y+1）<=2，即-2<=y<=0，

机值在边界达到

由已知条件可知，绝对值(x-1)<=2,即-1<=x<=3；绝对值2（y+1）<=2，即-2<=y<=0，

M=2x(x-1)+y^2，最小值M=2*3*（-2）+0=-12