Stationary and saddle points for $f(x,y) = x^3+x^2-xy+y^2+5$

Question

Find Stationary and saddle points for $f(x,y) = x^3+x^2-xy+y^2+5$

What I have tried: $$\begin{align}f_x &= 3x^2+2x-y \\f_y &= -x+2y \\ \implies x&=2y,y = x/2 \end{align}$$

Plugging in the values for y in $f_x$ I only find $x=0$ and so I presume the stationary points are $(0,0), (2y, x/2)$. As for the saddle point this can be found with $f_{xx}f_{yy}-f{xy}$

$$\begin{align}f_{xx} &= 6x+2 \\ f_{yy}&=2 \\ f_{xy} &= -1 \\ \implies f_{xx}f_{yy}-f_{xy} &= -12x-4 \end{align}$$

However, these don't seem right to me- where might I have gone wrong?

Answer 1

If, in the equation $3x^2+2x-y=0$, you replace $y$ by $\frac x2$, then it becomes $3x^2+\frac32x=0$, whose roots are $0$ and $-\frac12$. So, the points where the gradient is null are $(0,0)$ and $\left(-\frac12,-\frac14\right)$.

The Hessian of $f$ at a point $(x,y)$ is $\left[\begin{smallmatrix}6x+2&-1\\-1&2\end{smallmatrix}\right]$. At $(0,0)$, this is equal to $\left[\begin{smallmatrix}2&-1\\-1&2\end{smallmatrix}\right]$, whose trace and determinant are both greater than $0$. So, the Hessian has two eigenvalues greater than $0$, and therefore $f$ has a local minimum at $(0,0)$.

On the other hand, the Hessian at $\left(-\frac12,-\frac14\right)$ is $\left[\begin{smallmatrix}-1&-1\\-1&2\end{smallmatrix}\right]$, whose determinant is negative. So, the eigenvalues have opposite signs, and therefore it's a saddle point.

Answer 2

Let $f(x,y)=x^{3}+x^{2}-xy+y^{2}+5$ and notice that $f\in \mathcal{C}^{+\infty}(\Omega\subseteq \mathbb{R}^{2}, \mathbb{R})$ then since $f$ is differentiable the critical points is the same as stationary points. A point $(x,y)\in \Omega$ is critical point if $\nabla f(x,y)=0$. Here we have $\nabla f(x,y)=(3x^{2}+2x-y,-x+2y)=0$ iff $(x,y)$ is $(-1/2,-1/4)$ or $(0,0)$.

• Hence the stationary points are $(-1/2,-1/4)$ and $(0,0)$.

The determinant of Hessian matrix for $f$ is given by $$\det Hf(x,y):=\det \begin{bmatrix} f_{xx}(x,y) & f_{xy}(x,y)\\ f_{yx}(x,y) & f_{yy}(x,y)\end{bmatrix}=\det \begin{bmatrix} 6x+2 & -1 \\ -1 & 2\end{bmatrix} $$

Remark:

A point $(x,y)\in \Omega$ is a saddle point if $\det Hf(x,y)<0$. A point $(x,y)\in \Omega$ is local minimum if $\det Hf(x,y)>0$ and $f_{xx}(x,y)>0$. Finally, a point $(x,y)\in \Omega$ is a local maximum if $\det Hf(x,y)>0$ and $f_{xx}(x,y)<0$.

Hence

• Since $Hf(-1/2,-1/4)=\det \begin{bmatrix}6\left(-\frac{1}{2}\right)+2 & -1\\ -1 & 2 \end{bmatrix}=-3<0$ so $(-1/2,-1/4)$ is a saddle point.
• Since $Hf(0,0)=\det \begin{bmatrix}6(0)+2 & -1\\ -1 & 2 \end{bmatrix}=3>0$ and since $f_{xx}(0,0)=2>0$ then $(0,0)$ is a local minimum.