求z=x^2+y^2+1在y=1-x下的极值
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求z=x^2+y^2+1在y=1-x下的极值
求z=x^2+y^2+1在y=1-x下的极值
求z=x^2+y^2+1在y=1-x下的极值
z=x^2+y^2+1
=x^2+(1-x)^2+1
=2x^2-2x+2
=2(x^2+x+1)
=2[x^2+2*1/2*x+(1/2)^2-(1/2)^2+1]
=2[(x+1/2)^2+3/4]
=2(x+1/2)^2+3/2
当=-1/2时,z最小为3/2
F=x^2+y^2+1+r(y-1+x)
Fx=2x+r=0
Fy=2y+r=0
y=1-x
解得:x=y=1/2为极值点
求z=x^2+y^2+1在y=1-x下的极值
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