# Finding local max/min and saddle points of $f(x,y)=x^2+xy+y^2+y$

I have a question asking to find the local max/min/saddle point of $f(x,y)=x^2+xy+y^2+y$

I have:

$f_x=2x+y$

$f_y=x+2y+1$

$f_{xx}=2$

$f_{yy}=2$

$f_{xy}=1$

$D=2*2-1^2=3$

I know if D is more than zero then there should be extrema, and I had found (1/5,-2/5), (1,-2) as critical points, but with no way to test critical points by plugging them into $f_{xx}$ since it's a constant.... I'm not really sure where to go from here. They''re simple derivatives and I don't see where I messed up on them anywhere so I'm not sure what else to try.

• Here I see $f_y = x + 2y + 1$. – Rolf Hoyer Oct 15 '16 at 4:06
• Yea, typo, sorry. That's what I have on my paper. – windy401 Oct 15 '16 at 4:08
• Just because $f_{xx}(x,y) =2$ doesn't mean we can't evaluate it at points! $f_{xx}(1/5, -2/5) = 2$ – erfink Oct 15 '16 at 4:25
• Yea but every point would then give the same result, fxx > 0, so local min. So every critical point is a local min? The book is just saying the answer is that there is a local min at $x=1/3$ and $y=-2/3$. Still trying to figure out how they solved for that critical point.... Unfortunately I'm not good at solving for the zeros needed....Sigh. – windy401 Oct 15 '16 at 4:48

To get the critical points we need to solve $(f_x,f_y)=(0,0)$ for $(x,y)$. This gives two equations: (1) $2x+y=0$ and (2) $x+2y+1=0$. (1) is equivalent to $y=-2x$, and plugging this into (2) gives $x+2(-2x)+1=0$, so $3x=1$ and hence $x=1/3$. But then $y=-2x=-2/3$, showing that $(1/3,-2/3)$ is the only critical point.
To see if this is a local max/min, you need to plug this into the Hessian, which is the (constant) matrix $$\begin{pmatrix}2&1\\1&2\end{pmatrix}.$$ Since this matrix is positive definite we have a local min. at $(1/3,-2/3)$.