extrema problem

x^2+y^2+z^2=0 ???????

2011-05-16 23:02:32 補充：
x+y+z=0, z=-(x+y)
x^2+y^2+(x+y)^2=6 , x^2+xy+y^2= 3
(x+ y/2)^2+ (3/4)y^2=3
Set x+ y/2= √3 cost, y=2sint, then
x=√3 cost - sint, y=2 sint, z=-√3 cost- sint, -π<= t <= π
f(x,y,z)= 2x+yz= 2√3 cost- 2sint+ 2sint(-√3 cost - sint)=g(t)
g(t)=2√3 cost -2sint-√3 sin(2t)+cos(2t) -1
g'(t)= -2√3 sint -2cost- 2√3 cos(2t) - 2sin(2t)
= -2√3 [sint+ 1-2(sint)^2] -2[ cost+ 2sint cost]
= 2√3(sint-1)(2sint+1)- 2cost(1+2sint)]
=-2(1+2sint)[cost+√3 (1-sint)]
=-2(1+2sint)√(1-sint) [√(1+sint) +√3 √(1-sint)]
g'(t)=0, t= π/2, -π/6, -5π/6
g(-π/6)= 5 is the maximum value
g(π/2)=g(-5π/6)= -4 is the minimum value