Question

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?

Answer 1

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