已知x、y、z为非负有理数,且满足3x+2y+z=5,x+y-z=2.设t+2x+y-z,求t的最大值,最小值各是多少?【提示：把x、y、z中任意一个字母作为独立变数或常数.】

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t = 2x+y-z = 2x+(y-z) = 2x+(2-x) = 2+x ≥2+0 = 2t = 2+x ≤ 2+5/3 = 11/3综上,t...