高考问答北京音乐餐厅名字大全:一道初二数学题
来源：百度文库 编辑：高考问答 时间：2024/05/11 03:26:13
已知z、x、y三个非负数，满足2x+y-3z=1，3x+z+2y=5,若记v=y-7z+3x,试求v的最大值和最小值。
2x+y-3z=1 => y = 3z - 2x + 1
3x+z+2y=5 => z = 5 - 3x - 2y
=> x = 7z -3
y = 7 -11z
=> v = v=y-7z+3x = 3z - 2
z > = 0
=> v > = -2
又因3x+z+2y=5，x，y，z为非负数，所以z=<5
=> v<=3*5-2=13
所以v的最大值为13，最小值为-2
2x+y-3z=1 => y = 3z - 2x + 1
3x+z+2y=5 => z = 5 - 3x - 2y
=> x = 7z -3
y = 7 -11z
=>v = v=y-7z+3x = 3z - 2
z > = 0
=> v > = -2
所以v的最小值为－2，最大值为正无穷。
2x+y-3z=1 => y = 3z - 2x + 1
3x+z+2y=5 => z = 5 - 3x - 2y
=> x = 7z -3
y = 7 -11z
=> v = v=y-7z+3x = 3z - 2
z > = 0
=> v > = -2
又因3x+z+2y=5，x，y，z为非负数，所以z=<5
=> v<=3*5-2=13
所以v的最大值为13，最小值为-2