(Continuing with the perimeter of 60 and units omitted.)

As I have returned to this question many times over the years, it seems to me remarkable how the way of denoting the sides is presented as if it is “canonical” — that is, call one side x and the other 30-x [because it is the Right Thing To Do].

Now you have a function: Area(x) = x(30-x) = 30x – x^2.

Since x > 0, this is a downward facing parabola; i.e., its maximum occurs at the vertex.

Calculus? d/dx 30x – x^2 = 30 – 2x, which is 0 when x=15.

Algebra II? Vertex is at x = -b/2a = -30/2(-1) = 15.

Etc.

The suggestion from earlier is *not* to pick the same way of labeling sides always used, i.e., *not* to pick x and 30-x. If you want, you can think of the suggested approach as a substitution: let n=15-x, which means x=15-n.

Now adjacent sides are 15-n and 15+n.

So, this time, Area(n) = 225 – n^2.

Only “sense-making” (if forced to classify by topic, I would call this Real Analysis) is needed to realize that the area is now maximized when n=0.

Thus, adjacent sides are 15-0 and 15+0, i.e., 15×15 square.

From the standpoint of integrated algebra and geometry, this “substitution” is shifting the parabolic function that represents the area by 15, so that it is now symmetric about the origin.

I’m not sure why the problem’s history involves this marriage between x and 30-x; I find parabolas generally *much* easier to deal with when they are symmetric about the origin, i.e., when the coefficient of x is 0.

In fact, this idea of shifting — which is a frequent if confusing topic at the (graphical) intersection of algebra and geometry — is how I consider questions about quadratics more generally, e.g., how to find the x coordinate of the vertex and how to uncover the quadratic equation; cf. http://matheducators.stackexchange.com/a/9709

Since n^2 is non-negative, subtracting it off will either lower the total area (uh oh) or leave it alone (iff n=0). The latter case means we oughtn’t vary the 15 at all. That is, the square yields maximal area.

OMG. All of the light bulbs in my head.

1) If phrased in its (contrived-ish) standard form of, given a fixed amount of fencing — create a rectangular enclosure that maximizes interior area, then there are some very (I think) interesting variations.

Fixing numbers for concreteness: Suppose you have 60m of fencing. What are the dimensions of the rectangular enclosure that maximizes interior area … if you also have access to a very long wall? (So, now you only need to worry about building 3 sides of the enclosure.) Once you find the answer, does it jibe well with your intuition? If so, why? If not, how can you update your thinking? (You can ask other questions, e.g., about what happens if you have access to a 10m wall. And for something like Desmos, what if you have access to an N meter wall, where you can vary N? Etc.)

2) With regard to, “I finally found some use for this fact that takes up a significant chunk of my brain’s random access memory” … here is another way to think about the problem, which I didn’t see (may have missed?) in the post and comments above. (I *bet* it could be implemented in Desmos in some interesting way…)

When you have a rectangle with (say) perimeter 60m, any pair of adjacent sides add up to 30m. (Why?) So, omitting units, let’s just *start* by using the square, i.e., all four sides have length 15, and the area is 15×15 = 225.

I think that this maximizes the enclosed area. If you think otherwise, then you will need to change some side; but if you change 15 to, say, 15+n, then maintaining the perimeter requires changing its adjacent side(s) to 15-n.

Then the area becomes (15+n)(15-n) = 225 – n^2 …

Since n^2 is non-negative, subtracting it off will either lower the total area (uh oh) or leave it alone (iff n=0). The latter case means we oughtn’t vary the 15 at all. That is, the square yields maximal area.

The above constitutes the underlying mathematics, and *not* a “lesson plan” … but, I think it is an interesting way of thinking about the problem!

(Going beyond: The alternative *idea* in my latter comment is ess’ly a geometric interpretation of the AM-GM Inequality. This needn’t be broached explicitly when presenting it to students, but knowing this piece of background info can/could be helpful in situating the “fact” within a more general mathematical body of knowledge related to inequalities and means, which have applications in finance and elsewhere.)

This year I want to let students design ‘dining tables’ that have a desired number of people around it (all of the kids in the class), and let them find the ‘table’ that would have the most area for food. Then after they have come up with lots of designs, they can actually cut out a model of their tables and pin them to a graph.

Love that image.

https://drive.google.com/open?id=0B-5elry4t9OOOEMtLVRqWVQ0WEk

I did this with purely rectangles when I did this graph, but would love to extend it to something like a table that they can actually create. As a fun product of the lesson I then want students to stand up and actually make a couple of the tables to see that it really is the one with the biggest area, like this teacher did with 6th graders here at around 9min if you don’t wanna sift through the whole video.

https://www.teachingchannel.org/videos/real-world-geometry-lesson

The purpose of using this problem for me is to introduce the idea of quadratics giving you the ability to maximize a quantity, and also to introduce why there is symmetry in a parabola, because you can have a 3×4 table or a 4×3 table and they give you the same area.