若x+y+z=30,3x+y-z=50,x,y,z均为非负数,求m=5x+4y+2z的取值范围
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若x+y+z=30,3x+y-z=50,x,y,z均为非负数,求m=5x+4y+2z的取值范围
若x+y+z=30,3x+y-z=50,x,y,z均为非负数,求m=5x+4y+2z的取值范围
若x+y+z=30,3x+y-z=50,x,y,z均为非负数,求m=5x+4y+2z的取值范围
130
x+y+z=30 (1)
3x+y-z=50 (2)
(1)+(2) 得 4x+2y=80
即 2x+y=40
因此 y=40-2x
(2)-(1)得 2x-2z=20
即 x-z=10
因此 z=x-10
所以 M=5x+4y+2z=5x+4(40-2x)+2(x-10)=140-x
所以M的最大值当x最小时取得
又因为x,y,z非负
又由 x-z=10 z大于等于0 所以x大于等于10
因此x最小为10
所以m的最大值为140-10=130
5x+4y+2z=2*(x+y+z)+(3x+y-z)+z=2*30+50+z=110+z
因为z>=0
所以m>=110
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