# What are the extrema and saddle points of f(x,y) = x^2+xy+y^2+y?

Dec 28, 2016

I found no saddle points, but there was a minimum:

$f \left(\frac{1}{3} , - \frac{2}{3}\right) = - \frac{1}{3}$

To find the extrema, take the partial derivative with respect to $x$ and $y$ to see if both partial derivatives can simultaneously equal $0$.

${\left(\frac{\partial f}{\partial x}\right)}_{y} = 2 x + y$

${\left(\frac{\partial f}{\partial y}\right)}_{x} = x + 2 y + 1$

If they simultaneously must equal $0$, they form a system of equations:

$2 \left(2 x + y + 0 = 0\right)$
$x + 2 y + 1 = 0$

This linear system of equations, when subtracted to cancel out $y$, gives:

$3 x - 1 = 0 \implies \textcolor{g r e e n}{x = \frac{1}{3}}$

$\implies 2 \left(\frac{1}{3}\right) + y = 0$

$\implies \textcolor{g r e e n}{y = - \frac{2}{3}}$

Since the equations were linear, there was only one critical point, and thus only one extremum. The second derivative will tell us whether it was a maximum or minimum.

${\left(\frac{{\partial}^{2} f}{\partial {x}^{2}}\right)}_{y} = {\left(\frac{{\partial}^{2} f}{\partial {y}^{2}}\right)}_{x} = 2$

These second partials are in agreement, so the graph is concave up, along the $x$ and $y$ axes.

The value of $f \left(x , y\right)$ at the critical point is (by plugging back into the original equation):

$\textcolor{g r e e n}{f \left(\frac{1}{3} , - \frac{2}{3}\right)} = {\left(\frac{1}{3}\right)}^{2} + \left(\frac{1}{3}\right) \left(- \frac{2}{3}\right) + {\left(- \frac{2}{3}\right)}^{2} + \left(- \frac{2}{3}\right)$

$= \frac{1}{9} - \frac{2}{9} + \frac{4}{9} - \frac{6}{9} = \textcolor{g r e e n}{- \frac{1}{3}}$

Thus, we have a minimum of $\textcolor{b l u e}{f \left(\frac{1}{3} , - \frac{2}{3}\right) = - \frac{1}{3}}$.

Now, for the cross-derivatives to check for any saddle points that could be along a diagonal direction:

${\left(\frac{{\partial}^{2} f}{\partial x \partial y}\right)}_{y , x} = {\left(\frac{{\partial}^{2} f}{\partial y \partial x}\right)}_{x , y} = 1$

Since these are both in agreement as well, instead of being opposite signs, there is no saddle point.

We can see how this graph looks just to check: