实数x、y满足x2+y2=4,则x+y-xy的最大值为_.
问题描述:
实数x、y满足x2+y2=4,则x+y-xy的最大值为______.
答
∵实数x、y满足x2+y2=4,
∴可设x=2cosθ,y=2sinθ.
令t=sinθ+cosθ=
sin(θ+
2
)(θ∈[0,2π)),π 4
∴t∈[−
,
2
].
2
则t2=1+2sinθcosθ,可得2sinθcosθ=t2-1.
∴x+y-xy=2cosθ+2sinθ-4sinθcosθ
=2t-2(t2-1)
=−2(t−
)2+1 2
≤5 2
,5 2
当且仅当t=
时,x+y-xy取得最大值为1 2
.5 2
故答案为:
.5 2