设x∈R+且x2+y2/2=1,求x√1+y2的最大值

设x∈R+且x2+y2/2=1,求x√1+y2的最大值

∵(x^2+y^2)/2=1,∴x^2+y^2=2
x√(1+y^2)= √[x^2(1+y^2)
≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2
∴最大值为3/2

设x=sinα（α∈(0,π)），y=√2cosα
则，x√(1+y^2)=√[x^2(1+y^2)]
=√[sin^2 α*(1+2cos^2 α)]
=√[sin^2 α*(3-2sin^2 α)]
令sin^2 α=t∈(0,1]，则原式=√[t*(3-2t)]
=√(-2t^2+3t)
f(t)=-2t^2+3t=(-2)[t^2-(3/2)...

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设x=sinα（α∈(0,π)），y=√2cosα
则，x√(1+y^2)=√[x^2(1+y^2)]
=√[sin^2 α*(1+2cos^2 α)]
=√[sin^2 α*(3-2sin^2 α)]
令sin^2 α=t∈(0,1]，则原式=√[t*(3-2t)]
=√(-2t^2+3t)
f(t)=-2t^2+3t=(-2)[t^2-(3/2)t]=(-2)[t^2-(3/2)t+(9/16)]+(9/8)
=(-2)[t-(3/4)]^2+(9/8)
因为t∈(0,1]，所以f(t)|max=f(3/4)=9/8
所以，原式的最大值=√(9/8)=(3√2)/4

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