f ( x , y ) = 4 + x 3 + y 3 − 3 x y f(x,y)= 4+x^3 + y^3 -3xy f ( x , y ) = 4 + x 3 + y 3 − 3 x y
f x = 3 x 2 − 3 y f_x=3x^2-3y f x = 3 x 2 − 3 y
f y = 3 y 2 − 3 x f_y=3y^2-3x f y = 3 y 2 − 3 x
Find the critical point(s)
f x = 0 f_x=0 f x = 0 f y = 0 f_y=0 f y = 0 Then
3 x 2 − 3 y = 0 3x^2-3y=0 3 x 2 − 3 y = 0 3 y 2 − 3 x = 0 3y^2-3x=0 3 y 2 − 3 x = 0
y = x 2 y=x^2 y = x 2
x 4 − x = 0 x^4-x=0 x 4 − x = 0
Critical points are ( 0 , 0 ) , ( 1 , 1 ) (0, 0), (1, 1) ( 0 , 0 ) , ( 1 , 1 )
f x x = 6 x f_{xx}=6x f xx = 6 x
f x y = f y x = − 3 f_{xy}=f_{yx}=-3 f x y = f y x = − 3
f y y = 6 y f_{yy}=6y f yy = 6 y
D = ∣ f x x f x y f y x f y y ∣ = ∣ 6 x − 3 − 3 6 y ∣ = 36 x y − 9 D=\begin{vmatrix}
f_{xx} & f_{xy} \\
f_{yx} & f_{yy}
\end{vmatrix}=\begin{vmatrix}
6x & -3 \\
-3 & 6y
\end{vmatrix}=36xy-9 D = ∣ ∣ f xx f y x f x y f yy ∣ ∣ = ∣ ∣ 6 x − 3 − 3 6 y ∣ ∣ = 36 x y − 9 Point ( 0 , 0 ) (0,0) ( 0 , 0 )
D ( 0 , 0 ) = 36 ( 0 ) ( 0 ) − 9 = − 9 < 0 D(0,0)=36(0)(0)-9=-9<0 D ( 0 , 0 ) = 36 ( 0 ) ( 0 ) − 9 = − 9 < 0 Then f ( 0 , 0 ) f(0,0) f ( 0 , 0 ) is not a local maximum or minimum. Point ( 0 , 0 ) (0,0) ( 0 , 0 ) is a saddle point of f . f. f .
Point ( 1 , 1 ) (1,1) ( 1 , 1 )
D ( 1 , 1 ) = 36 ( 1 ) ( 1 ) − 9 = 27 > 0 D(1,1)=36(1)(1)-9=27>0 D ( 1 , 1 ) = 36 ( 1 ) ( 1 ) − 9 = 27 > 0
f x x ( 1 , 1 ) = 6 ( 1 ) = 6 > 0 f_{xx}(1,1)=6(1)=6>0 f xx ( 1 , 1 ) = 6 ( 1 ) = 6 > 0
Then f ( 1 , 1 ) = 3 f(1,1)=3 f ( 1 , 1 ) = 3 is a local minimum.
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