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

由y=1/4x7/4-->1/4x-7/4 x>=-1
y=1/4x-7/4-->1/4x-7/4x z>=-7x/4-7/4--> z>=-7(*3)/4-7/4=-7
因此有 -7=亲 我那个答案上写的Z的最大值是0能再算一下么 会追加分的嗯，后面的判断错了点,更正如下：由y<=x-1, y>=1/4x-7/4-->1/4x-7/4<=x-1--> x>=-1y<=-2x+5, y>=1/4x-7/4-->1/4x-7/4<=-2x+5-->x<=3,因此有： -1= y=z+2xy<=x-1-->z+2x<=x-1-->z<=-x-1--> z<=1-1=0y<=-2x+5--> z+2x<=-2x+5--> z<=-4x+5--> z<=4+5=9y>=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<=0, z<=9, z>=-7所以只能取三者的交集：-7=