丨x-1丨十|x+2丨十丨8-y丨十丨y+1丨=12,其中ⅹ+y的最大值和最小值为多少

2022-11-05 23:25发布

12=丨x-1丨十|x+2丨十丨8-y丨十丨y+1丨={2x+2y-6,x≥1,y≥8;{2y-4,-2
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1楼 · 2022-11-05 23:38.采纳回答

12=丨x-1丨十|x+2丨十丨8-y丨十丨y+1丨
={2x+2y-6,x≥1,y≥8;
{2y-4,-2<x<1,y≥8;
{-2x+2y-8,x≤-2,y≥8;
{2x+10,x≥1,-1<y<8;
{12,-2<x<1,-1<y<8;
{-2x+8,x≤-2,-1<y<8;
{2x-2y+6,x≥1,y≤-1;
{10-2y,-2<x<1,y≤-1;
{6-2x-2y,x≤-2,y≤-1.
由上述第一、第五、第九段函数知,x=1,y=8时x+y取最大值9;x=-2,y=-1时x+y取最小值-3.