Question

已知x、y為正實數,且2x+8y-xy=0,求x+y的最小值.

Answer 1

解:由2x+8y-xy=0,得2x+8y=xy.

∴+=1.

∴x+y=(x+y)(+)

=10++

=10+2(+)

≥10+2×2×

=18.

    當且僅當=,即x=2y時,取等號.

    又2x+8y-xy=0,

∴x=12,y=6.

∴當x=12,y=6時,x+y取最小值18.