## Problem on classifying critical points

Find the critical points of $f(x,y) = 2 x^2 -2 x^2 y + y^2$. Classify them as minima, maxima, or saddle points.
• ## Solution

#### Finding the critical points

Recall that
We now compute the values of $x$ and $y$ where $\partial_x f=0$ and $\partial_y f=0$.
The critical points are then given by any $x,y$ that solve \begin{align} 4x - 4 x y &=0 \\- 2 x^2 + 2y &= 0 \end{align}
which simplifies to \begin{align} x(1-y) &=0 \tag{1}\\ y &= x^2 \tag{2} \end{align}
Equation (1) gives us two cases: (a) $x=0$ or (b) $y=1$.

(a) In the case $x=0$, equation (2) gives us that $y=0$. Hence, $(0,0)$ is a critical point.
(b) In the case $y=1$, equation (2) gives us that $x^2 = 1$. Thus, $x$ can be $1$ or $-1$. We see that $(1,1)$ and $(-1,1)$ are also critical points.

We have found all the critical points: $(0,0)$, $(1,1)$, and $(-1,1)$.

#### Classifying the critical points

In order to classify these points as maxima, minima, or saddle points, we recall the second derivative test:
In order to analyze the three critical points, we must first compute $D$. We compute the second derivatives of $f$ as:
\begin{align}
\partial_{x x} f &= 4 -4 y \\
\partial_{x y} f = \partial_{y x} f&= -4 x \\
\partial_{y y} f &= 2
\end{align}
For $(x,y) = (0,0)$, we see $D = 8 > 0$ and $\partial_{x x} f > 0$. Hence $(0,0)$ is a minimium.
For $(x,y) = (1,1)$, we see $D = -16 < 0$. Hence $(1,1)$ is a saddle point.
For $(x,y) = (-1,1)$, we see $D = -16 < 0$. Hence $(-1,1)$ is a saddle point.
The critical points are $(0,0)$ (minimizer), $(1,1)$ (saddle point), and $(-1,1)$ (saddle point).