# 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.

## 2 Comments »

1. The problem statement is incomplete. There is no restriction stated under which the maximum area is thought. Also, the typos in the text are inexcusable for a mathematician/teacher who puts an elementary piece for a public view.

Comment by Anatoli — June 15, 2013 @ 3:38 pm

• This was done quickly and informally; I expect that interested readers would be able to fill in details.

But since you seem upset, I’ll refund the money you spent to read this. 🙂

Comment by blueollie — June 15, 2013 @ 4:17 pm