Question

实数x、y满足方程x^2+2y^2-2xy+x-3y+1=0,则y的最大值为?

Answer 1

x^2+2y^2-2xy+x-3y+1=0

(x-y)^2+(y-1)^2+x-y=0

(x-y)(x-y+1)+(y-1)^2=0

令y-1=a,y=a+1

（x-a-1）(x-a)+a^2=0

x^2-(2a+1)x+2a^2+a=0

△=（2a+1）^2-4(2a^2+a)=0

4a^2+4a+1-8a^2-4a=0

-4a^2+1=0

a=±1/2

y最大值是1+1/2=3/2

y最小值是1-1/2=1/2