Question

x+3y+2z=1,求3x^2-y^2+2z^2的最小值

Answer 1

解：：令u=3x^2-y^2+2z^2,将x=1-3y-2z代入，
那么26y^2-2(18z-9)y+(14z^2-12z-u-3)=0
△ =4(18z-9)^2-4*26(14z^2-12z-u-3)≥0
即40z^2+12z-(3+26u)≤0
那么△=12^2-4*40(3+26u) ≥0
所以u ≥-3/20
所以当x=-1/20,y=9/10,z=-3/20时，取最小值u=-3/20.

Answer 2

u = 3x^2-y^2+2z^2 - k (x+3y+2z)
u'(x) = 6x - k = 0
u'(y) = -2y - 3k = 0
u'(z) = 4z - 2k = 0
所以x = k/6 ，y = -3k/2 ，z = k/2
代入x+3y+2z = 1得k/6-3k/2+k/2 = 1所以k = -6/5
所以当x = -1/5，y=9/5，z=-3/5的时候取得最小值
3/25 - 81/25 + 18/25 = -60/25 = -12/5