3D Maxima and Minima

In multivariable calculus, we were finding relative maxima and minima. It’s much like finding maxima and minima in 2D.

The general idea in 2D is that if you go a little bit to the left or a little bit to the right (changing x by a wee bit) at a maxima or minima, you aren’t really changing your height much (you aren’t changing y by much). Another way to look at it… if you zoom in enough to a maxima or minima, you’ll almost see a straight line! And you can make it as “straight” as you want it by zooming in more and more and more.

Does that make sense?

Now we do a similar argument for maxima and minima in 3D:

At the top of peaks or troughs, you’ll notice if you walk a wee little bit in the x direction, the height (z) isn’t changing by much. Similarly if you move a wee little bit in the y direction, the height isn’t changing as much. (Or, analogously, if you zoom in a lot lot lot lot, you’ll be looking at something almost perfectly flat, a horizontal plane…)

In other words, instead of saying maxima and minima only occur when f'(x)=0, we now can say that maxima and minima only occur when f_x(x,y)=0 and f_y(x,y)=0. That’s the mathematical way to talk about moving a bit in a x-direction or y-direction.

So my kids know to find possible relative maxima or minima, you have to find the points (x,y) which make f_x(x,y)=0 and f_y(x,y)=0.

In class I then posed a few good questions:

(a) If you know a maximum occurs at the point (2,3), how can you show that the directional derivative in the direction <\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}> is also 0?

(b) If you know that at the point (5,7), the directional derivative for <\frac{3}{5},\frac{4}{5}> is 0, and the directional derivative for <\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}> is also 0. Prove that the point (5,7) is a maximum or minimum or saddle point.

These were things that I thought up on the fly… it’s interesting. We get so used to procedures, that we sometimes forget what they mean. The point I was trying to make is that if any two (different) directional derivatives were 0 at a point, then that point could be a maxima or minima. If you pose that as a claim, and students are used to thinking algebraically, they have to go through the motions to see this is true. (It basically involves creating and solving a system of two linear equations…). [1]

But there’s a much easier way to get students to buy that claim. If you think graphically, this makes sense… if you are at a maxima or minima, and you zoom in enough, the surface will look like a flat plane. So of course if you walk a short distance in any direction, you shouldn’t be moving (much) in the z direction.

I don’t know… this isn’t deep or anything. But it was something that I didn’t plan in class that I thought was interesting…

[1] This is how it would go… Assume you know D_{<a,b>}f=0 and D_{<c,d>}f=0 (and the two vectors aren’t scalar multiples of each other). Then you can rewrite D_{<a,b>}f=af_x+bf_y=0 and D_{<c,d>}f=cf_x+df_y=0. Well then you simply have a system of equations that you can solve for f_x and f_y — and it is easy enough to show that the solution is f_x=0 and f_y=0.

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4 comments

  1. This is cool! Nice way of explaining it…

    Any thought as to the assumptions that make all of this work? I have discontinuous counterexamples in mind (where all of the directional derivatives exist at a point but these properties don’t hold). Eg f= x on the x-axis, f = y on the y-axis, f=0 everywhere else.

  2. Oh yeah, the functions need to be “nice.” Specifically what “nice” means would require me to do some thinking on my snowday… or look something up in a book… both things I’m prolly not going to do. But you know, standard nice things. :)

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