Question

已知点P（X,Y）在直线x+y-4=0上,O是原点,求OP最小值

Answer 1

y=4-x

OP^2=x^2+y^2=x^2+(4-x)^2=2x^2-8x+16

=2(x-2)^2+8

当x=2,则y=2时,OP的最小值是:根号8=2根号2