$f(x,y)=x^2 +xy + y^2 +y$
I find the critical point $p=\left(\frac{1}{3},-\frac{2}{3}\right)$ by solving the system of partial equations.
I plug this ordered pair into the function and get $f(p) = -\frac{1}{3}$.
I thought my next step was to compute $f_{xx}(p)$ and $f_{yy}(p)$ and $f_{xy}(p)$ and evaluate $D$.
$f_{xx}(p)=2$ and $f_{yy}(p)=2$ and $f_{xy}(p)=1$
So, $D>0$ and $f(p) <0$ which seems to indicate a local max. Book says minimum. Confused.
$\endgroup$ 11 Answer
$\begingroup$$Hf(x,y)= \begin{pmatrix}f_{xx} & f_{xy}\\ f_{yx} & f_{yy}\\\end{pmatrix}$
$Hf(\frac{1}{3},\frac{-2}{3})=\begin{pmatrix}2 & 1\\ 1 & 2\\\end{pmatrix}$
$\det(Hf(\frac{1}{3},\frac{-2}{3})) = 3 > 0$
The second derivative test for two variables states that:
If at the critical point $f_{xx} > 0$ and $\det(Hf(x,y)) > 0$ then at the critical point you have a local minimum.
Otherwise if at the critical point $f_{xx} < 0$ and $\det(Hf(x,y)) > 0$ then at the critical point you have a local maximum.
Otherwise if at the critical point $\det(Hf(x,y)) < 0$ then at the critical point you have a saddle point.
Otherwise if at the critical point $\det(Hf(x,y)) = 0$ then the test is inconclusive.
Thus for your function at $(\frac{1}{3},\frac{-2}{3})$ you have a local minimum.
