x^2+2xy+2y^2+2y=0,x+2y最大值

x^2+2xy+2y^2+2y=0,x+2y最大值

条件可整理为(x+y)²+(y+1)² = 1.
于是(x+2y+1)² = (x+y)²+(y+1)²+2(x+y)(y+1) ≤ 2((x+y)²+(y+1)²) = 2.
得x+2y+1 ≤ √2,即x+2y ≤ √2-1.
可验证当x = 1,y = √2/2-1时等号成立,故x+2y的最大值为√2-1.