设实数x,y≥0,且满足2x+y=5,则函数f(x,y)=x²+xy+2x+2y的最大值是?
答案是49/4
人气:320 ℃ 时间:2019-08-21 20:46:00
解答
x,y≥0且2x+y=5,
y=5-2x≥0,x≤5/2
故f(x,y)=x²+xy+2x+2y=x²+x(5-2x)+2x+2(5-2x)
=-x²+3x+10
=-(x-3/2)²+49/4
≤49/4
当且仅当x=3/2时取得,此时y=5-2x=2
故f(x,y)最大值为49/4
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