已知x,y,z为非负实数,且满足x+y+z=30,3x+y-z=50.求u=5x+4y+2z的最大值和最小值.
人气:457 ℃ 时间:2019-08-21 12:20:39
解答
将已知的两个等式联立成方程组x+y+z=30①3x+y−z=50②,所以①+②得,4x+2y=80,y=40-2x.将y=40-2x代入①可解得,z=x-10.因为y,z均为非负实数,所以40−2x≥0x−10≥0,解得10≤x≤20.于是,u=5x+4y+2z=5x+4(...
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