Question

x,y,z为三个非负有理数,且满足3x+2y+z=5,x+y-z=2,若2x+y...

x,y,z为三个非负有理数,且满足3x+2y+z=5,x+y-z=2,若2x+y-z=s,则s的取值范围是什么?

Answer 1

已知
x+y-z
=
2
所以
y-z
=
2-x
因为
x是非负有理数
所以
x
≥0
又因为
3x＋2y＋z＝5
所以当y
=
0
,z
=
0时
x有最大值为
xmax
=
5/3
则
s
=
2x+y-z
=
2x+(y-z)
=
2x+(2-x)
=
2+x
≥2+0
=
2
s
=
2+x
≤
2+5/3
=
11/3
综上,s的最大值为
smax
=
11/3
s的最小值为
smin
=
2
s的取值范围是[2,11/3]