已知z=x^2+5/4y^2-xy-x-1/2y+1,则z的最小值是
问题描述：
已知z=x^2+5/4y^2-xy-x-1/2y+1,则z的最小值是
答
z=x^2+5/4y^2-xy-x-1/2y+1
=x^2-(y+1)x+5/4y^2-1/2y+1
=[x-(y+1)/2]^2-(y+1)^2/4+5/4y^2-1/2y+1
=[x-(y+1)/2]^2+y^2-y+3/4
=[x-(y+1)/2]^2+(y-1/2)^2+1/2
所以 z 的最小值是1/2,当且仅当 x=3/4,y=1/2时取等