已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值

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已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值已知x,y属于R+,且2x+8y-xy=0,求x

已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值
已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.
2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值

已知x,y属于R+,且2x+8y-xy=0,求x+y的最小值.2.已知x,y属于R+,且x+2y=3,求[1/(x+2)]+[1/2(y+1)]的最小值
利用重要不等式的性质
x,y>0,2x+8y=xy则2/y + 8/x =1则x+y=(x+y)(2/y + 8/x )
=2x/y +8y/x +10
> =8+10=18(均值不等式)
(当2x/y=8y/x即x=12,y=6时取=)
2)
x+2y=3
(x+2)+2y+2=7
[1/(x+2)]+[1/2(y+1)]
=[1/(x+2)]+[1/2(y+1)]*[(x+2)+2y+2]/7
=1/7 [1+2(y+1)/(x+2)+(x+2)/2(y+1)+1]
=1/7[2+2(y+1)/(x+2)+(x+2)/2(y+1)]
>=1/7[2+2根号1]
=4/7
当取得=时
2(y+1)=x+2

1、2x+8y-xy=0,所以y=2x/(x-8)
x+y=x+2x/(x-8)
=x+(2(x-8)+16)/(x-8)
=x+2+16/(x-8)
=x-8+16/(x-8)+10
因为x>0,y>0。即2x/(x-8)>0
所以x-8>0
则原式≥2√((x-8)16/(x-8))+10=18,此时x-8=16/(x-8),x=12

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1、2x+8y-xy=0,所以y=2x/(x-8)
x+y=x+2x/(x-8)
=x+(2(x-8)+16)/(x-8)
=x+2+16/(x-8)
=x-8+16/(x-8)+10
因为x>0,y>0。即2x/(x-8)>0
所以x-8>0
则原式≥2√((x-8)16/(x-8))+10=18,此时x-8=16/(x-8),x=12
最小值为18
2、x+2y=3,所以2y=3-x
[1/(x+2)]+[1/2(y+1)]
=1/(x+2)+1/(3-x+2)
=1/(x+2)+1/(5-x)
=7/((x+2)(5-x))
((x+2)(5-x)≤(x+2+5-x)²/4=49/4
所以原式≥4/7,当x+2=5-x,即x=3/2时取等号
所以最小值为4/7

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