# Testing critical points using newton-raphson method

I'm working on a homework assignment for my artificial intelligence course and I became a bit stuck on one part.

The problem asks to retrieve the optima of the function and then to tell for each optima whether it is a minimum or a maximum.

The function is as follows: $$f(x,y,z) = x \ln x + y \ln y + z \ln z + \alpha(x + y + z - 1)$$

The gradient was calculated to be: $$\nabla f(x,y,z) = (\ln x + \alpha + 1)\vec{i} + (\ln y + \alpha + 1)\vec{j} + (\ln z + \alpha + 1)\vec{k}$$

In order to find the critical points I then set $\nabla f(x,y,z) = 0$, which gave me the following critical points:

$$(0,0,0),(\frac{1}{e^{\alpha + 1}},\frac{1}{e^{\alpha + 1}},\frac{1}{e^{\alpha + 1}}),(1,1,1)$$

What is the best approach to test if each critical point it is a maximum or a minimum in the function?

I also have calculated the Hessian matrix for an earlier part of this problem. The hessian matrix came to be:

$$\left[ \begin{array}{ccc} \frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z} \end{array} \right]$$

Is there a way to utilize the hessian matrix to test for convexity at those points and therefore if convex it is a minimum? The problem with that thought was that when I calculated it out they all turned out to be minimums which doesn't seem right to me.

• Try graphing the function. Use Wolfram Alpha, for example. A picture is worth ... – bubba Feb 2 '14 at 2:27
• You've missed 24 stationary points. – Jeff Snider Feb 2 '14 at 3:55
• Can you explain how you came to 24? – dmcqu314 Feb 2 '14 at 3:58
• Minima are when the Hessian is Positive Definite. Maxima = Negative Definite. Very easy to check for a diagonal matrix. – Jeff Snider Feb 2 '14 at 3:59
• @bubba this is a 4d equation it didn't graph too nicely. Or do you mean to graph the gradient vector? – dmcqu314 Feb 2 '14 at 4:01

This is tricky around 0. Here is how I would approach it:

The Jacobian is $$\nabla f(x,y,z)=(\ln x+\alpha +1)i+(\ln y+\alpha +1)j+(\ln z+\alpha +1)k.$$

First note that $x,y,z=0$ is not in the domain, but this function has a value in the limit: $$\lim_{x\to0^+} (x\ln x+\alpha x)=0.$$ We treat points where some or all of $x,y,z=0$ as though they are in the domain, because to exclude them means there is no optimum near $(0,0,0)$ although the function is increasing as $x,y,z\to 0^+$ and bounded above in the local region. This means that $(0,0,0)$ is a local maximum: any point with $x,y,z<0$ does not exist, and any nearby point with $x,y,z>0$ takes a smaller value.

We have stationary points for all $x,y,z\in\{0,e^{-\alpha-1}\}$. This means we have $2^3=8$ stationary points to consider.

The Hessian is $$\left(\begin{array}{ccc}\frac{1}{x}&0&0\\0&\frac{1}{y}&0\\0&0&\frac{1}{z}\end{array}\right).$$

We have determined we should treat $(0,0,0)$ as a local maximizer. The Hessian is undefined here, but at a local maximizer it must be negative definite. For a coordinate with a zero value (e.g., $x=0$) we treat that row of the Hessian as though it contains a negative value on the diagonal and is zero elsewhere.

Next consider $(e^{-\alpha-1},e^{-\alpha-1},e^{-\alpha-1})$. At this point the Hessian has positive values along the diagonal and is zero elsewhere: it is positive definite. Thus this point is a local minimizer.

Consider the remaining stationary points, where at least one coordinate is $0$ and at least one coordinate is $e^{-\alpha-1}$. Here the Hessian has a positive value in diagonal in the row corresponding to $e^{-\alpha-1}$ and an effectively negative value in the row corresponding to $0$. Hence these are all saddle points.

Since the function is unbounded above as $x,y,z\to\infty$ there is no global maximizer. But the function is bounded below, hence $(e^{-\alpha-1},e^{-\alpha-1},e^{-\alpha-1})$ is the global minimizer.

• Ah that makes a lot more sense now. Thanks! – dmcqu314 Feb 2 '14 at 18:40