| (1) f(x,y) =5x^2+5xy+3y^2 =5(x+y/2)^2+7y^2/4≧0
(x,y)=(0,0) のときf(x,y)の最小値 0
(2) f(x,y) =-4x^3+18xy-9y^2
f_x=-12x^2+18y=0 -2x^2+3y=0 f_y=18x-18y=0 x-y=0 x=y -2x^2+3x=0 x(3-2x)=0 x=0.or.x=3/2 (x,y)=(0,0).or.(x,y)=(3/2,3/2)
f_xx=-24x f_xy=18 f_yy=-18
(f_xx)(f_yy)-(f_xy)^2=18*24x-18^2 (x,y)=(0,0)のとき (f_xx)(f_yy)-(f_xy)^2=-18^2<0だから極値でない
(x,y)=(3/2,3/2)のとき (f_xx)(f_yy)-(f_xy)^2=18*36-18^2>0 f_xx=-24*3/2=-36<0 極大値 f(3/2,3/2) =-4(3/2)^3+18(3/2)^2-9(3/2)^2 =27/4
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