### Or maybe they just studied really hard.

My early afternoon precaclulus class has always been a shade weaker than my late afternoon precalculus class; class averages tend to differ by around 5% from one class to the next.

Until this week. Class average on the early afternoon test: 58%. I found this bewildering, as I thought that this test was considerably easier than the previous two. My late afternoon class agreed with me: their class average was 78%.

I don’t like to assume the worst of my students, but no mathematician can look at the data I’m looking at, without incorporatinng their knowledge of standard deviations and variance and accuracy within such and such percent so many times out of twenty and concluding that in all likelihood, something is awry. Since my two classes are one right after the other, with only the ten minute break to get between buildings, I’d naively assumed that it would be okay to give the same test to both sections. Curiously, my early afternoon class - the ones who bombed the test - tended to leave the classroom early. Most were gone an hour into the ~~sixty~~ *eighty* (thanks, rohan!) minute test, which left them with half an hour to talk to the late afternoon students. By contrast, I had to pry the test out of half of my late afternoon students’ fingers.

Next week, I’ll write two tests. This shouldn’t be much extra work, as many of my students are so weak that “find the maximum possible product of two numbers that sum to 50″ is a completely different question from “find the maximum possible product of two numbers that sum to 60″. But this still leaves the question of what to do this time. I hope that at the very least, the early afternoon informants were being paid well for their sacrifice.

Could you tell what method you taught to solve the problem “find the maximum possible product of two numbers that sum to 50″? For fun, I tried to solve it, and the only solution that was immediately apparent to me, in the few minutes I worked on the problem, was to employ a quick bit of Calculus. The product of the two numbers would be x(50-x), or 50x-(x^2); I set that quation equal to zero, took the derivative of it and solved for x, which turns out to be the intuitive value of 25. So the answer is 25x25, or 625. But I thought you didn’t teach a Calculus class, so I’m curious how your students would’ve been expected to solve this.

to be the devil’s advocate here : what if you are a better teacher the second time around? also, early afternoon = right after lunch = a bad time for many students to focus.

The times I have had two sections of a class, covering the same material, I have unfortunately found that the second time through I do a better job. A significantly better job. I have also observed that for me, early morning and just-after-lunch seminars are disasters. Around 10-1 and after 4, partial differential equations, Sobolev embedding theorems and chern classes suddenly make a lot more sense. Presumably this is true for completing the square, graphing functions and factoring polynomials.

[Wes : complete the square. you don’t need calculus to observe that -y^2 + C <= C and attains this maximum at y=0]

I’ve always been intrigued by problems like that; it seems obvious that the answer is 25, but actually proving it is a bit more trouble. The way I learnt is to first differentiate the equation once and set it to zero to find all the minima/maxima, then differentiate

again(if more than one minima/maxima exists) and test the concavity/convexity of the curve around the extreme to ‘prove’ that it is a maximum. But surely there should be a non-Calculus way? :-)(by the way, did you mean to say an hour into the

ninetyminute test? :-)Sam -

to be the devil’s advocate here : what if you are a better teacher the second time around? also, early afternoon = right after lunch = a bad time for many students to focus.Oh, for sure - I am certainly better the second time around. Also, I just plain happen to have more very strong students in the later class, and more very weak students in the earlier class. (I also completely messed up a lecture on inverse functions in the early class, and nailed it the second time around - but that accounts for maybe a 5% difference) However, as I said - the fact that

all of a suddenI have a 20% gap in marks, where previously it was under 10%, is probably non-random.Also - did you just say that Chern classes were making more sense? WHERE THE HELL WERE YOU TWO YEARS AGO? And could you explain them to me? I’m still convinced that they don’t really exist.

Completing Sam’s comment…

if you complete the square, to get the quadratic in the form -(a - x)^2 + C

You know that the smallest the square term can be is 0, so the largest the whole thing could be is C.

Anyway, that’s how one finds the max/min for a quadratic, without calculus.

My 8am class is so totally dominating the 12:20 class. Far and away. But it varies: 8am section that filled up during “regular registration” = pretty good students. 8am section that was created a few days before the start of the semester = awful.

Were I explaining that problem to my College Algebra students of Fall 2003, I would have them draw the graph and find the vertex.

I returned the tests the other day, and felt a bit better about things. Although some of the disparity can be attributed to the fact that I’m just doing a (slightly) better job teaching the later class than the earlier clsas, the later class is just stronger and more motivated - a difference that gets magnified during the term, as the units I covered on the test relied on the previous units, which relied on the previous unit…

Also, the typical student in my earlier class isn’t much weaker than the typical student in my late afternoon class. However, my earlier class has a lot of very weak students (ones who can’t add fractions, ones who have no idea how to handle word problems), and my later class has far more very strong students.

We’ll see how test #4 and the exam go.

My solution: By symmetry (x\to 50-x), if there is a unique maximum, it must be at x=25. And this is a precalc test, so since you ask for a maximum, it must exist and be unique, by testmanship.

Better argument: Just look at parabola, and you can tell that it has a unique maximum, and is symmetrical, so that maximum must be at x=25.

Beyond that, I tend to appeal to naive calculus: consider a rectangle, sides x and 50-x. If I add a thin strip to the long side and take a thin strip off the short side, then I have a net gain, so certainly no non-square can be the maximum. On the other hand, area is a cont’s function of x, and x=0 and x=50 both give less than x=10, so there must be some maximum, and the square is the only possible one.

Meep&Sam’s soln is on the face more elementary; certainly it doesn’t have to appeal to naive calculus, etc., but I’d say it’s much less intuitive. This is partly because I find geometry almost always more understandable than algebra, and partly because limiting arguments and naive calculus are very important skills to develop for real life. Very few people actually need to be able to differentiate rigorously and all, but I can think of myriad applications of naive calculus (in business, econ, psych, etc., and heck, in all of physics).

Heh, I didn’t mean for this to be a symposium on maximizing quantities described by quadratic equations. Theo - I agree that the symmetry argument is more intuitive than the completing the square method that I showed in class, but keep in mind that half of my students don’t find fractions intuitive :-/

I’m just glad that I know Calculus so that I can easily solve problems like maximizing quantities without resorting to extremely tricky and advanced techniques like visualization of symmetry, completing the square and fractions. ;)