# College Math Teaching

## April 3, 2010

### Optimization Problems in Calculus: symmetries in solutions

Filed under: calculus, optimization, student learning — collegemathteaching @ 9:23 pm

For reasons I’ve yet to comprehend, “business calculus” students often have trouble with the following kind of problem:

Suppose you wish to fence in a rectangular plot of land. You have one long wall to use for one side. The sides perpendicular to the wall cost 10 dollars per foot of fencing and the side parallel to the long wall costs 30 dollars per foot of fencing. What is the largest area that you can enclose?

Yes, there are those who can’t even figure out that the area is denoted by $A= xy$ and some have trouble with the cost function $C = 20x + 30Y$

Some struggle with the algebra.

But let’s look at the mathematics itself: there is a certain symmetry to the solution; if one views the cost function as $C = Ax + BY$ and the area function $V = xy$ then one can show that the solution $x=x_0,y=y_0$ will always have the following property: $Ax_0 = By_0 = (1/2)C$

In fact it is a pleasant exercise to show that the solution to the following problem:
Maximize $x^{r}y^{s}=V (r\geq 1, s\geq 1)$
Subject to $Ax+By = C, A > 0, B > 0, C > 0$
is given by $x = (C/A)(r/(r+s)), y = (C/B)(s/(r+s))$

That is, $Ax = C(r/(r+s)), By =C(s/(r+s))$ or that the “share” of the cost associated with the $Ax$ term depends on the fraction $r/(r+s))$, which is the fraction of the total exponent of the function determining $V$.

If we look at the associated dual problem (minimize the cost given that we need the area $V$) we find that there is a more complicated symmetry at hand:
$x = ((rb)/(sa))^{s/(r+s)}V^{1/(r+s)}, y = ((sa)/(rb))^{r/r+s}V^{1/(r+s)}$.
Note the symmetry: if we take the ratio of the solution components we obtain:
$x/y = (r/s)(B/A)$.
How cool is that? 🙂
Note: I know that this follows immediately from the Lagrange Multipliers technique; still it is fun to derive via single variable calculus techniques.

### Popular Media Article on Exponentials and Logarithms

Filed under: calculus, mathematics education, media — collegemathteaching @ 8:50 pm

Power Tools By STEVEN STROGATZ

[…]
But when you need mathematical dynamite, it’s time to unpack the exponential functions. They describe all sorts of explosive growth, from nuclear chain reactions to the proliferation of bacteria in a Petri dish. The most familiar example is the function 10x, in which 10 is raised to the power x. Make sure not to confuse this with the earlier power functions. Here the exponent (the power x) is a variable, and the base (the number 10) is a constant — whereas in a power function like x2, it’s the other way around. This switch makes a huge difference. Exponential growth is almost unimaginably rapid.

That’s why it’s so hard to fold a piece of paper in half more than 7 or 8 times. Each folding approximately doubles the thickness of the wad, causing it to grow exponentially. Meanwhile, the wad’s length shrinks in half every time, and thus decreases exponentially fast. For a standard sheet of notebook paper, after 7 folds the wad becomes thicker than it is long, so it can’t be folded again. It’s not a matter of the folder’s strength; for a sheet to be considered legitimately folded n times, the resulting wad is required to have 2n layers in a straight line, and this can’t happen if the wad is thicker than it is long. […]

The rest of the article is worth reading; I’ll probably send it to my “business calculus” students.

## April 1, 2010

### Portability of skills and optimization problems

Filed under: calculus, student learning — collegemathteaching @ 3:25 pm

Consider the following optimization problem: you have a long wall and you wish to put a rectangular plot next to the wall; the wall forms a barrier for one of the 4 sides:

Say that the fencing for the sides perpendicular to the wall costs 3 dollars per foot and the side parallel to the wall costs 10 dollars per foot.

Naturally, there are two questions you could ask:
1. Given that the area must be, say, 1000 square feet, what is the cheapest fence that you can build or
2. Given that you have only 420 dollars to spend, how big of an area can you enclose?

Neither of these questions is difficult, but then I am teaching this to a class which has a median math ACT of 22 (off semester for this course). This fact is relevant for what follows.

I did problem number 2 first:

Objective: Maximize $A = xy$
Constraint: $C= 6x + 10y = 420$
(of course, $x > 0, y > 0$)

Then we proceed as follows: use the constraint to eliminate a variable:
$420 - 6x = 10y$ which gives $42 - (6/10)x = y$
Now substitute into the objective: $A = x(42 - (6/10)x)$
$A =42x -(6/10)x^2$
Differentiate the objective and set equal to zero:
$42-(12/10)x =0$ which has solution $x =35$ and $y = 21$.

No biggie, right? Well, pay attention to the step in which we solved for $y$.

Now when we did the dual problem, I had the students “help me out”.

Objective: minimize $C= 6x + 10y$
Constraint $xy = 1000$

So what did one student do? You guessed it: $y = 1000 -x$
Yes, this student had passed their algebra class.

What gives? Why didn’t the student simply say $y = 1000/x$?

After seeing a few more answers like this it dawned on me: on the first problem they saw me subtract, so they figured that subtraction was the step you used to go from the constraint to the objective!

Had I just asked them “solve for $y$ if $xy=1000$ without having the optimization problem to do, they could have done it.

In short, their algebra skills are not yet “portable”; they don’t see the “constraint to the one variable objective function” as merely an application of the algebra that they already “know”.

Sometimes it is tough to see where a bad student gets lost, given that most of us who teach calculus were strong students when we first took the course.