Day 15 in Mat329

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Today:

Announcements
- You have an assignment due today.
- There's an exam coming up next Thursday. It will cover through 14.6. I'll start off Tuesday with your questions.
- Lunch! Right after class. Run up there and eat. You can even eat twice, if you've already been. Go back for dessert.
Last time we began section 14.7: maximum and minimum values. Today we'll do a few application problems.
Consider the following problem: find all the local maxima and minima (and saddle points) of the function $f(x,y)=x^2+y^2+x^2y+4$.
As a polynomial, this has degree 3 (since we have an $x^2y$ term). This means that it is not restricted to the bowl or hyperbolic paraboloid of quadratic functions, but may have more interesting features. On the other hand, we should realize that, provided the function is twice differentiable at any point, it can be approximated by a "tangent bowl" there.
I've used the term "tangent bowl", but it could just as well be a "tangent saddle" (because that's another shape that a quadratic function of two variables can have). Let's take a look at these two:

So we expect the local behavior to have one of these two shapes (if we go beyond the tangent plane, to the tangent bowl/saddle/quadratic shape).
So how are we to determine the shape of $f(x,y)$, and discover the location of its extrema? The extrema at differentiable points will look like bowls or saddles, locally and generically.
There are other possibilities, however: for example

These were summarized well in a short note that my dad published in the American Mathematical Monthly, long, long ago. I'd almost forgotten it, but thought to recreate it using Mathematica.
The univariate case is handled by considering the places where the derivative is equal to 0, and a very similar situation arises in the multivariate case: the analogy is that the gradient will equal the 0 vector.
The argument is simple, given on p. 970: let's go over it. Once again we rely on our understanding of univariate functions to arrive at this conclusion.

For the function $f(x,y)$ given above, we will find that there were two saddles, and a minimum. The graph of $f$ suggests it (notice the contour plots):

Now, how do we know that these are the only critical points, their locations, etc.? We compute the gradient vector, and set it equal to zero. This gives us a pair of equations to solve simultaneously for the critical points:
- $f_x(x,y)=2x+2xy$
- $f_y(x,y)=2y+x^2$
Setting these to zero simultaneously yields two non-linear equations in two unknowns (this is often a difficult problem to solve); when we solve them in this case (using substitution), we obtain the three solutions
1. $(x,y)=(0,0)$
2. $(x,y)=(\sqrt{2},-1)$
3. $(x,y)=(-\sqrt{2},-1)$
Now how can we determine whether we have a minimum or a maximum or a saddle at a given critical point? If the second partials are continuous on a disk with center $(a,b)$, a critical point, then define \[ D = D(a,b) = f_{xx}(a,b) f_{yy}(a,b) - [f_{xy}(a,b)]^2. \] There are three cases:
1. If $D>0$ and $f_{xx}(a,b)>0$, then $f(a,b)$ is a local minimum.
2. If $D>0$ and $f_{xx}(a,b)<0$, then $f(a,b)$ is a local maximum.
3. If $D<0$, then $f(a,b)$ is neither a local maximum nor minimum (in this case, $(a,b)$ is called a saddle point.
So now we compute $D$ in this case:
- $f_{xx}(x,y)=2+2y$
- $f_{xy}(x,y)=2x$
- $f_{yx}(x,y)=2x$ (ah ha! we should have known better: this is a polynomial, after all....)
- $f_{yy}(x,y)=2$
Hence $D=4(1+y)-4x^2$. In two cases, $D$ is negative, and we have a saddle; in the other case, at the origin, $D=4$ and $f_{xx}=2>0$, so we have a local minimum.
Notice that this is a natural extension of the univariate second derivative test. If we know that we have a bowl or an umbrella, then all we have to do is take a cross-section through the bottom of the bowl or top of the umbrella, and apply the second derivative test to the univariate cross section. If the second derivative is positive, it's a bowl; if the second derivative is negative, it's an umbrella.
If we're interested in absolute extrema, then we need to consider the boundary of a region, as well. There is an extreme value theorem for multivariate functions -- a natural extension of the EVT for univariate functions (p. 975).
Now let's take a look at some examples. I want to start with one of the most important, however: #55, p. 979.
- #50
- #52

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