#### A Math Forum Project

 ESCOT Problem of the Week: Archive of Problems, Submissions, & Commentary

Student Version

### Llama III - posted December 15, 1999

(Try Llama I and Llama II before you begin this problem.)

You and Lester decide to go into business building llama pens, since you had such great results with Latisha's pen. You want to create some general building guidelines to help in your business, and need to know:

Given any length of fence, what proportion of height to width will create a pen with the largest possible area?

Experiment with three different lengths of fencing and find the ideal height/width ratio for 36-, 25-, and 50-meter lengths of fencing. Write your building guidelines for any length of fence in the pad. (Don't forget to explain how you arrived at your building guidelines.)

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Sad to say, we only had one submission to Llama III, possibly because of winter vacation. (Or maybe llamas aren't Y2K compliant?) A sample correct solution is posted below. A few students had a sense of the solution to this part of the problem in Llama I. Some students knew that the pen with height 9m and width 18 m would work because 18 = 2 * 9, but none of these students explained why they thought this.

Happy New Year!

Sample Solutions

First question: (proportion of h to w)

For all three fences the ratio of height to width reduces to 1/2, so I would tell Lester that the ideal proportion of height to width in order to create the pen with the largest area is 1:2.

Second Question: (experiment with lengths)

36m: I started with the 36-meter fence and remembered that the best dimensions were height 9 meters, and width 18 meters.

25m: Then I went to the situation with 25 meters of fence and found that the largest area I could get was 68 which happened in two places, where the fence had dimensions 6.5 by 12 meters, and where it had dimensions 6m by 13m. Since I don't think there can be two largest areas,(or because the curve from last week was symmetric about the maximum or some idea to that effect) I tried taking the average of 6 and 6.5, and got a height of 6.25m, the average of the two widths is (12 + 13)/2, which is 12.5m (so h=6.25 and w=12.5). The area of this pen (6.25m by 12.5 m) is 78.125sq meters, which is the maximum. (Students may confirm this a bunch of ways, but I think it makes sense through symmetry, they also may try other values on either side, or graph it on a calculator.)

50m: For the third pen with 50 meters of fence, I experimented and discovered one maximum area where the fence is 12.5m by 25m. The maximum area is 312.50 square meters. I know this is the maximum because a fence 12m by 26m has the same area as a fence 13m by 24m, both having the area of 312 square meters.