Question

已知实数x，y满足方程组y≤x-1，y≤-2x+5，y≥1/4x-7/4 求z=y-2x的最大值

Answer 1

由y<=-1, y>=1/4x7/4-->1/4x-7/4<=x-1--> x>=-1
y<=-2x+5, y>=1/4x-7/4-->1/4x-7/4<=-2x+5-->x<=3
因此有： -1=<x<=3
由z=y-2x--> y=z+2x
y<=x-1--> z+2x<=x-1-->z<=-x-1--> z<=1-1=0
y<=-2x+5--> z+2x<=-2x+5--> z<=-4x+5--> z<=4+5=9
y>=1/4x-7/4--> z+2x>=1/4x-7/4--> z>=-7x/4-7/4--> z>=-7(*3)/4-7/4=-7
因此有 -7=<z<=9
Z最大为9.

追问

亲 我那个答案上写的Z的最大值是0 能再算一下么 会追加分的

追答

嗯，后面的判断错了点,更正如下：
由y=1/4x-7/4-->1/4x-7/4 x>=-1
y=1/4x-7/4-->1/4x-7/4x y=z+2x
y z+2xz z z+2x z z=1/4x-7/4--> z+2x>=1/4x-7/4--> z>=-7x/4-7/4--> z>=-7(*3)/4-7/4=-7
即有:z=-7
所以只能取三者的交集：-7=<z<=0
故Z最大值为0.