若x>=0,y>=0,且x+y=1,则x^2+y^2的最大值是_________.
热心网友
最简方法: ∵x^2+y^2=(x+y)^2-2xy=1-2xy ∴当x或y等于0时取得最大值1
热心网友
令x=(cosT)^2,y=(sinT)^2,则x=(1+cos2T)/2,y=(1-cosT)/2x^2+y^2=(1+cos2T)^2/4+(1-cos2T)^2/4=[1+(cos2T)^2]/2 & (cos2T)^2=1+(cos2T)=x^2+y^2=<1.
热心网友
解:y=1-x∴x^+y^=2x^-2x^+1=2(x-1/2)^+1/2∵x≥0,y=1-x≥0∴0≤x≤1,即:-1/2≤x-1/2≤1/2,∴0≤(x-1/2)^≤1/4∴当x=0或x=1时,x^+y^取最大值1.或解:设x=cos^θ,y=sin^θ,∴x^+y^=(cos^θ)^+(sin^θ)^=1-(1/2)sin^2θ≤1.[注:sin^2θ=0如θ=nπ/2时"="成立]
热心网友
换元，得1