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问题: 求z中的最大值和最小值

设z=2x+y,使式中变量x,y满足条件x-4y<=-3;3x+5y<=25;x>=1

解答:

z=(7/17)(x-4y)+(9/17)(3x+5y)<=(7/17)(-3)+(9/17)(25)=12
z=(9/4)x-(1/4)(x-4y) >=(9/4)(1)-(1/4)(-3) =3
3<=z<=12, 最大值 =12, 最小值 =3