求二元函数f(x,y)=x平方+y平方+xy,在条件x+2y=4下的极值

x+2y=4
x=4-2y 代入方程得
f(4-2y,y)=(4-2y)^2+y^2+y(4-2y)
=16-16y+4y^2+y^2+4y-2y^2
=3y^2-12y+16
=3(y^2-4y)+16
=3(y^2-4y+4)+16-12
=3(y-2)^2+4
当y=2 x=4-2y=0时有极值 f(x,y)=4