Question

如果实数x、y满足等式x^2+y^2=2x+4y,求2x-y的最小值

Answer 1

F(x,y)=2x-y+m(x^2+y^2-2x-4y)

F`x=2+m(2x-2)=0

m=-2/(2x-2)=1/(1-x)

F`y=-1+m(2y-4)=0

m=1/(2y-4)

1-x=2y-4

x=5-2y

(5-2y)^2+y^2=2(5-2y)+4y

y^2-4y+3=0

y=1 y=3

x=3 x=-1

2x-y=5 or -5

2x-y的最小值-5