Find the minimum value of the function f(x,y)=8x^2+6y^2+20xy

和刚才方法相同,受限函数为:2x^2+5xy=1
设函数F(x,y)=8x^2+6y^2+20xy+4x+5y+4 +λ(2x^2+5xy-1)
∂F/∂x=16x+20y+4+4λx+5λy=0,
λ=-(16x+20y+4)/(4x+5y),(1)
∂F/∂y=12y+20x+5+5λx=0,
λ=-(20x+12y+5)/(5x),(2)
联立(1)和(2)式,
85y^2+48xy=0,
y=0,
2x^2+0=1,
y=0,x=±√2/2,
y=-48x/85,
x^2=-17/14,无实数解,
∴x=±√2/2,y=0,选x=-√2/2有最小值,
∴f(x,y)(min)=8*(-√2/2)^2+6*0+20*0+4*(-√2/2)+5*0+4
=8-2√2.