求z=x^2+y^2+1在y=1-x下的极值_数学解答

求z=x^2+y^2+1在y=1-x下的极值

人气：208 ℃　时间：2020-07-11 10:31:26

解答

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