Question

若a^2+b^2=1,求a+b的最大值

Answer 1

你好

a^2+b^2≥2ab
a^2+b^2=1
2ab≤1
（a+b）^2=a^2+b^2+2ab≤2
a+b的最大值是√2

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Answer 2

因为(sinx)^2+(cosx)^2=1
所以可令a=sinx,b=cosx
所以a+b=sinx+cosx=√2sin(x+π/4)
所以a+b的最大值为√2

Answer 3

令x= a+b
b=x-a
代入
2a²-2ax+(x²-1)=0
a是实数则△>=0
4x²-8x²+8>=0
x²<=2
-√2<=x<=√2
所以最大值是√2

Answer 4

let
a= cosy
b = siny
S = a+b
= cosy +siny
S' = -siny +cosy=0
tany =1
y = π/4
S''(π/4)<0 (max)
max S = S(π/4) = √2

Answer 5

a^2+b^2=1,设a=cosx,b=sinx ,x∈[0,2π)则
a+b=cosx+sinx=√2sin(x+π/4)≤√2