已知{2x+y-5>=0,3x-y-5<=0,x-2y+5>=0},则(x+1)^2+(y+1)^2的最大值和最小值为?

2025-05-05 01:20:48
推荐回答(2个)
回答1:

2x+y-5>=0 (1)
0>= 3x-y-5 (2)
x-2y+5>=0 (3)
(3)*3+(2)=>
3x-6y+15>=3x-y-5
=>y<=4
(1)*3+(2)*2=>
6x+3y-15>=6x-2y-10
=>y>=1
(2)*2+(3)=>
x-2y+5>=6x-2y-10
=>x<=3
(1)*2+(3)=>
x>=1
=> 1<=x<=3 1<=y<=4
(1)=>x>=(5-y)/2
=>(x+1)^2+(y+1)^2>=[(7-y)/2]^2+(y+1)^2=5y^2/4-6y/4+53/4=5(y-3/5)^2/4-9/5+53/4
当y=1时,值最小=13
(2)=> x<=(5+y)/3
=>(x+1)^2+(y+1)^2<=[(8+y)/3]^2+(y+1)^2
当y=4时,值最大=41

回答2:

1≤x≤3
3≤y≤4
20≤(x+1)^2+(y+1)^2≤41