Question

若实数x，y满足x^2+y^2+xy=1，则x+y的最大值是多少？

Answer 1

三角换元法，根据三角函数的有界性求出范围
x^2+y^2+xy=(x+y/2)^2+(3/4)*y^2=1
令x+y/2=cosθ ,(√ 3/2)y=sinθ
∴x+y=cosθ+(√ 3/3)sinθ=(2√3/3)sin(θ+π/3),
∴(x+y)max=2√3/3

Answer 2

令a=x+y
y=a-x
则x²+a²-2ax+x²+ax-x²=1
x²-ax+(a²-1)=0
x是实数则判别式△>=0
a²-4a²+4>=0
a²<=4/3
-2√3/3<=a<=2√3/3
所以最大值是2√3/3