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=2
所以当y = 0 ,z = 0时 x有最大值为 = 2/3
则
s = 2x+y-z = 2x+(y-z) = 2x+(2-x) = 2+x ≥2+0 = 2
s = 2+x ≤ 2+2/3 = 8/3
综上,
s的最大值为 smax = 8/3
s的最小值为 smin = 2
S的最大值与最小值之和是2+8/3=14/3
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=2
所以当y = 0 ,z = 0时 x有最大值为 = 2/3
则
s = 2x+y-z = 2x+(y-z) = 2x+(2-x) = 2+x ≥2+0 = 2
s = 2+x ≤ 2+2/3 = 8/3
综上,
s的最大值为 smax = 8/3
s的最小值为 smin = 2
S的最大值与最小值之和是2+8/3=14/3