Question

设正实数x,y,z满足x²-3xy+4y²-z=0,则当取得最大值时,x+2y-z的最大值为(　　)

(A)0     (B)      (C)2     (D)

Answer

C

【解析】由题得z+3xy=x2+4y2≥4xy(x,y,z>0),

即z≥xy,≥1.当且仅当x=2y时等号成立,

则x+2y-z=2y+2y-(4y2-6y2+4y2)

=4y-2y2=-2(y2-2y)

=-2[(y-1)2-1]=-2(y-1)2+2.

当y=1时,x+2y-z有最大值2.故选C.