Farmer Ted Goes 3D

Abstract

In his recent paper entitled Ted Goes Natural [2], Greg Martin discusses the plight of Farmer Ted, who wants to minimize the cost of chicken wire to enclose a rectangular coop with a base of 190 square feet. However, Farmer Ted has difficulty purchasing 4,j56 feet of chicken wire because he is only able to purchase wire in integer lengths. After much deliberation, he decides to build an 11 foot x 17 foot coop (187 square feet), which has the best cost-efficiency of any possible coop with integer side lengths and area 190 square feet. Here we add another dimension to the story of Farmer Ted. In recent years, Farmer Ted has helped numerous neighbors efficiently build chicken coops. Then one local farmer asked a question that intrigued him: The farmer wants to build an animal cage with a volume of up to 50 cubic feet. What is the most cost-effective way to do this with integer side lengths? Farmer Ted had begun to think in 3D. Before analyzing the three-dimensional problem, we should briefly review some of Martin's work. He starts with the basic calculus optimization problem of finding the minimum perimeter of a rectangle given a fixed area. As many students know, the optimal shape is a square. Martin then offers this variation: Given a positive integer N, what are the dimensions of the rectangle with integer side lengths and area at most N whose area-to-perimeter ratio is largest among all such rectangles? In order to solve this problem, Martin makes the following definitions: Let s(n) = min (c d) where cd =n and c, d E N. Equivalently, s(n) = min (d nld) where d I n. This gives the minimum semiperimeter, or half of the actual perimeter, of a rectangle with area n with integer sides. Define F (n) = n /s (n), which is the maximum ratio of area to semiperimeter for a given n. (Using this expression, which is twice as large as the maximum ratio of area to perimeter, is more aesthetically pleasing and does not change the analysis.) Farmer Ted prefers to enclose 187 ft.2 rather than 190 ft.2 because F(187) = 187/28 > F(190) = 190/29. This leads Martin to identify the set