Let $f(x,y) = x^2y+y^2+xy$ be a function, I want to find its local extrema an minima.

I easily find that $f$ has 2 critical points: $(x,y)=(0,0)$ and $(x,y) = (-1,0)$.

In order to find its local extrema, I now find the Hessian matrix of the quadratic form: $H_f(a) = \begin{bmatrix}2y&2x+1\\2x+1&2\end{bmatrix}$

Therefore for $(x,y) = (0,0)$ I find $H_f(0,0) = \begin{bmatrix}0&1\\1&2\end{bmatrix}$ and $H_f(-1,0) =\begin{bmatrix}0&-1\\-1&2\end{bmatrix}$

I now have to determine wether those matrices are positive definite, semi-positive definite, semi-negative definite, negative definite or non-definite. In other term if $x \in \mathbb{R^2}$ I have to determine the sign of $x^T.H_f(a).x$

  • $\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}0&1\\1&2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = 2(x+y)^2$ Therefore $H_f(0,0)$ is definite positive, (0,0) is a local extremum.

  • $\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}0&-1\\-1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = -(x-y)^2$ Therefore $H_f(-1,0)$ is definite negative$, (-1, 0) is a local minimum.

Are my calculations correct?

  • $\begingroup$ Your calculation of $H_f(a)$ is incorrect - $\frac{\partial^2f}{\partial x^2}=2y$ $\endgroup$ – uniquesolution May 23 '16 at 8:19
  • $\begingroup$ Once you correct your partial derivative with respect to $ \ x \ $ and fix the $ \ f_{xx} \ $ entry, you'll find there is a third critical point and that all three are simple to characterize. $\endgroup$ – colormegone May 23 '16 at 8:49
  • $\begingroup$ I've edited my post $\endgroup$ – aribaldi May 23 '16 at 12:06
  • $\begingroup$ @RecklessReckoner How is there a third critical point? $\endgroup$ – aribaldi May 23 '16 at 13:12

The critical points are points $a \in \mathbb R^2$ such that $\mathrm {grad}\, f(a) = 0$. Now $\mathrm{grad}\, f(x,y) = (2xy + y , x^2 + 2y + x)$. You need to solve $$\begin{cases}2xy + y = 0 \\x^2 + 2y + x = 0\end{cases} \iff \begin{cases}y(2x+ 1) = 0\\x(x + 1) + 2y = 0\end{cases} \iff \begin{cases}y= 0 \,\,\text{or}\,\, x = -\frac{1}{2}\\x(x+1) + 2y = 0\end{cases} \iff \ldots$$

Think you can take it from here? From this you should find the third critical point which is

$(x,y) = \left(-\frac{1}{2}, \frac{1}{8}\right)$


You missed a critical point. You have: $$f_x = 0 \iff 2xy+y = 0 \iff y(2x+1) = 0 \iff y = 0 \,\vee x = -\tfrac{1}{2}$$ From $y=0$ you already found the critical points $(0,0)$ and $(-1,0)$, substitution of $x = -\tfrac{1}{2}$ into $f_y = 0$ gives a third critical point: $\left(-\tfrac{1}{2},\tfrac{1}{8}\right)$.

For $(0,0)$, $\det H = -1 < 0$ so that is a saddle point; likewise for $(-1,0)$. But for $\left(-\tfrac{1}{2},\tfrac{1}{8}\right)$, you have $\det H > 0$ so $f$ attains an extremum, a minimum since $f_{xx} > 0$ there.

Reference: second order partial derivative test.

Also, how did you calculate $x^T H x$? For example:

$$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}0&1\\1&2\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}y\\x+2y\end{bmatrix} = xy+y(x+2y) = 2y(x+y)$$which is not the same as $2(x+y)^2$; it is indefinite (so the same conclusion as above applies: it corresponds to a saddle point).


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