Question

(1)已知实数x,y满足x+y-4=0,则x²+y²的最小值是__________;

(2)已知实数x,y满足(x+2)²+y²=1,则2x-y的最大值是__________.

Answer 1

解析:(1)令m=(x,y),n=(1,1).

∵|m·n|≤|m||n|,∴|x+y|≤,

即2(x²+y²)≥(x+y)²=16.∴x²+y²≥8,故x²+y²的最小值是8.

(2)令m=(x+2,y),n=(2,-1),2x-y=t.

由|m·n|≤|m||n|,得|2(x+2)-y|≤=,即|t+4|≤.

解得-4-≤t≤-4.故所求的最大值是-4.

答案:(1)8  (2)-4