若实数x,y满足约束条件2x+3y?5≤02x?y?5≤0x≥0,则函数z=|x+y+1|的最小值是(  )A.0B.4C.83D.7

2024-11-30 20:23:39
推荐回答(1个)
回答1:

解答:解:作出

2x+3y?5≤0
2x?y?5≤0
x≥0
可行域如图,
2x+3y?5=0
2x?y?5=0
,可得A(
5
2
,0)

2x+3y?5=0
x=0
,可得B(0,
5
3
),
2x?y?5=0
x=0
,可得C(0,-5).
A、B.C坐标代入z=|x+y+1|,分别为:
7
2
8
3
,4,
又z=|x+y+1|≥0,当x=0,y=-1时,z取得最小值0.z=|x+y+1|取可行域内的红线段MN时x+y+1=0.z都取得最小值0.
故选A.