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Critical points

A function $f(x,y)$ can have local maxima, local minima and/or saddle points. These points are also called critical points / stationary points.

To find critical points, you calculate $f_x(x,y)$ and $f_y(x,y)$. Then, all points for which both partial derivatives are zero, are the critical points. That is, a function $f(x,y)$ has a critical point (or stationary point) at $(a,b)$ when $f_x(a,b)=0$ and $f_y(a,b)=0$.

When you found the critical point, you may want to classify it. This can be done using the second derivative test:

Suppose a function $f(x,y)$ has a critical point at $(a,b)$.
Then we can calculate $D=D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-\left[f_{xy}(a,b)\right]^2$

If $D>0$ and $f_{xx}(a,b)>0$, then $(a,b)$ is a local minimum
If $D>0$ and $f_{xx}(a,b)<0$, then $(a,b)$ is a local maximum
If $D<0$, then $(a,b)$ is a saddle point
If $D=0$, then the test is inconclusive

Example 1

Question: find and classify the critical points of $f(x,y)=2x^2+2xy+3y^2-4y$

Solution: we calculate both partial derivatives and set them equal to zero: $f_x(x,y)=4x+2y$ and $f_y(x,y)=2x+6y-4$; so we get the system of equations $$\begin{cases}4x+2y=0\\2x+6y=4\end{cases}$$ which has the (only) solution $x=-\frac{2}{5},~y=\frac{4}{5}$.

Therefore, the (only) critical point of $f(x,y)$ is $\boxed{\left(-\frac{2}{5},\frac{4}{5}\right)}$.

The question was to find and classify the critical points, so we also have to perform the second derivative test. Let's calculate the second partial derivatives; they are $f_{xx}(x,y)=4$, $f_{yy}(x,y)=6$, $f_{xy}(x,y)=2$. Now we're pretty lucky, because those second partial derivatives don't even depend on $x$ and $y$ anymore. So we can easily calculate $$\boxed{D=f_{xx}f_{yy}-[f_{xy}]^2=4\cdot6-2^2=20>0}$$ We also have, $\boxed{f_{xx}(-\frac{2}{5},\frac{4}{5})=4>0}$.

Therefore, with the second derivative test, we classify the critical point ${\left(-\frac{2}{5},\frac{4}{5}\right)}$ as a local minimum.