Find the Saddle Points of $f(x,y) = x^3 +3xy^2-3x^2-3y^2+4$

Question

I started by finding the partial derivatives :

$$ f_x(x,y) = 3x^2+3y^2-6x,$$ $$f_y(x,y) = 6xy -6y$$ but I cant seem to find the critical points from here.

Answer 1

To find critical points, we solve:

$$\begin{cases} 3x^2 + 3y^2 - 6x &=& 0 \\ 6xy - 6y &=& 0 \end{cases}$$

From the second equation, we obtain $x=1$ or $y=0$.

Putting $x=1$ to the first equation gives $y=\pm1$.

Putting $y=0$ to the first equation gives $x=2$ or $x=0$.

By the Second partial derivative test:

$$H(x,y) = \begin{pmatrix}f_{xx}(x,y) &f_{xy}(x,y)\\f_{yx}(x,y) &f_{yy}(x,y)\end{pmatrix} = \begin{pmatrix}6x-6 & 6y\\6y &6x-6\end{pmatrix}$$

Then:

$$D(x,y) = \det(H(x,y)) = (6x-6)^2 - (6y)^2$$

• $D(1,-1) = 0^2 - (-6)^2 = -36$
• $D(1,1) = 0^2 - (6)^2 = -36$
• $D(0,0) = (-6)^2 - (0)^2 = 36$
• $D(2,0) = (6)^2 - (0)^2 = 36$

Thus $(1,-1)$ and $(1,1)$ are two saddle points.

(created using Geogebra)