Question

已知向量a=(x,2),b=(1,y),其中x大于0,y大于0,若a·b=4,则1/x+2/y的最小值为?

尽可能详细一些。

Answer 1

a=(x,2),b=(1,y),其中x大于0,y大于0,若a·b=4,

x+2y=4
1/x+2/y

=1/4 *(1/x+2/y)*4
=1/4*(1/x+2/y)(x+2y)
=1/4(1+2y/x+2x/y+4)
=1/4(5+2y/x+2x/y)
>=1/4(5+2*√2y/x*2x/y)
=1/4×9
=9/4
最小值=9/4

Answer 2

a.b=x+2y=4
S = 1/x+2/y
= 1/(4-2y) + 2/y
S' = 2/(4-2y)^2 - 2/y^2 =0
y^2-(4-2y)^2=0
3y^2-16y+16=0
(y-4)(3y-4)=0
y= 4(rejected) or 4/3
y''= 8/(4-2y)^3 +4/y^3
y''(4/3) >0 (min)
min S = S(4/3)
= 1/(4-8/3) + 3/(4/3)
= 3/5+9/4
= 57/20

Answer 3

=1/4 *(1/x+2/y)*4
=1/4*(1/x+2/y)(x+2y)
=1/4(1+2y/x+2x/y+4)
=1/4(5+2y/x+2x/y)
>=1/4(5+2*√2y/x*2x/y)
=1/4×9
=9/4