已知{2x+y-5>=0,3x-y-5=0},则（x+1)^2+(y+1)^2的最大值和最小值为?
问题描述：
已知{2x+y-5>=0,3x-y-5=0},则（x+1)^2+(y+1)^2的最大值和最小值为?
答
1≤x≤3
3≤y≤4
20≤（x+1)^2+(y+1)^2≤41
答
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
6x+3y-15>=6x-2y-10
=>y>=1
(2)*2+(3)=>
x-2y+5>=6x-2y-10
=>x
x>=1
=> 1=[(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(x+1)^2+(y+1)^2