设x+y=t,则：y=t-x
x^2+y^2-16
=x^2+(t-x)^2-16
=2x^2-2tx+t^2-16=0有解
△=(2t)^2-4*2*(t^2-16)=4t^2-8t^2+128=-4t^2+128≥0
4t^2≤128
t^2≤32
-4√2≤t≤4√2
x+y的最大值4√2,最小值-4√2