若x+3y+3=0,则根号里(x-1)^2+(y-1)^2的最小值为

2024-12-04 00:40:34
推荐回答(1个)
回答1:

x+3y+3=0
x=-3y-3
(x-1)^2+(y-1)^2=(-3y-3-1)^2+(y-1)^2=(3y+4)^2+(y-1)^2
=9y^2+24y+16+y^2-2y+1=10y^2+22y+17
=10(y^2+11/5y+(11/10)^2-(11/10)^2)+17
=10(y^2+11/5y+(11/10)^2)-121/10+17
=10(y+11/10)^2+49/10
y=-11/10时有最小值为√(49/10)=7√10/10