求二元函数f(x,y)=4x^2+3y^2-xy-20x-21y+100的极值

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求二元函数f(x,y)=4x^2+3y^2-xy-20x-21y+100的极值
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求二元函数f(x,y)=4x^2+3y^2-xy-20x-21y+100的极值
求二元函数f(x,y)=4x^2+3y^2-xy-20x-21y+100的极值

求二元函数f(x,y)=4x^2+3y^2-xy-20x-21y+100的极值
看成x的一元函数,配方的:
f(x)=4*[x-(y+20)/8]^2-4[(y+20)/8]^2+3y^2-21y+100
=4*[x-(y+20)/8]^2+47/16*(y-4)^2+28
所以当x=(y+20)/8,y=4即x=3,y=4时,f取得fmin=28