1.设F(x,y,z)=x²－6xy＋10y²－2yz－z²＋18
∂z/∂x=-[∂F/∂x]/[∂F/∂z]=(2x-6y)/(2y+2z)=0 ,x=3y
∂z/∂y=-[∂F/∂y]/[∂F/∂z]=(6x-20y+2z)/(2y+2z)=0 ,3x-10y+z=0
联立可得y=z,代入原方程y=±3=z,x=±9,即点(9,3,3)和点(-9,-3,-3)为驻点
故z在(9,3)取最大值=3,在(-9,-3)取最小值=-3
2.易知f(x,y)=x²-y²+C,代入f(1,1)=2,C=2,f(x,y)=x²-y²+2,f'x(x,y)=2x=0,f'y(x,y)=-2y=0,可知f(x,y)在区域内取得驻点(0,0),这时f(x,y)=0.再考察边界上的点,x²+y²/4=1,x²=1-y²/4,代入f(x,y)得f(y)=3-5y²/4,y∈[-2,2]
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