The 36-Foot Garden: Same Perimeter, Different Area?
Mr. Chen's fourth graders design a rectangular garden bed that uses exactly 36 feet of fencing. Four proposals go up: 15 ft × 3 ft, 12 ft × 6 ft, 10 ft × 8 ft, and 9 ft × 9 ft. One group argues for 15 × 3 "because 15 is the biggest number on the list." Another group claims all four designs enclose the same area, since the perimeter is the same. Mr. Chen has 36 one-foot lengths of string per group, inch grid paper, and one rule: the bed must be rectangular, use all the fencing, and have whole-foot side lengths. He wants the argument settled by evidence the students generate, not by him.
1. Compute the perimeter and area of each proposal, rank them by area, and evaluate both claims.
Read the worked answer
| Dimensions | Perimeter | Area |
|---|---|---|
| 15 × 3 | 2(15 + 3) = 36 | 45 ft² |
| 12 × 6 | 2(12 + 6) = 36 | 72 ft² |
| 10 × 8 | 2(10 + 8) = 36 | 80 ft² |
| 9 × 9 | 2(9 + 9) = 36 | 81 ft² |
Ranking: 9 × 9 (81) > 10 × 8 (80) > 12 × 6 (72) > 15 × 3 (45).
First claim is false: 15 × 3 has the largest single side but nearly the smallest area. Making one dimension large forces the other to be small, and area depends on the product, not either factor alone.
Second claim is false: all four have perimeter 36, but areas range from 45 to 81 ft². The square encloses the most area among rectangles with a fixed perimeter - visible here because the areas climb as the two sides get closer together. So the correct choice is 9 × 9.
2. Describe a concrete investigation with the string and grid paper that would settle the "same perimeter, same area" claim, and list the data students should record.
Read the worked answer
Each group takes the 36 one-foot lengths and forms every whole-foot rectangle it can, moving one foot from the length to the width each time. Sides must be whole feet, so the widths run 1 through 9 (since perimeter 36 means length + width = 18, and letting width exceed length just repeats the same rectangles).
| Width | Length | Area (ft²) |
|---|---|---|
| 1 | 17 | 17 |
| 2 | 16 | 32 |
| 3 | 15 | 45 |
| 4 | 14 | 56 |
| 5 | 13 | 65 |
| 6 | 12 | 72 |
| 7 | 11 | 77 |
| 8 | 10 | 80 |
| 9 | 9 | 81 |
Data to record: width, length, area, and a sketch on grid paper. Then plot area against width: the points rise, peak at width 9, and would fall symmetrically if students continued to width 17. The plot is the evidence that perimeter is held constant while area changes, which directly refutes the second claim. A follow-up question - "what would happen if we were allowed decimals?" - lets students see that 9 × 9 is a maximum, not just the best of the whole-foot options.
3. Extend the problem: the bed will sit flat against the school wall, so fencing is needed on only three sides, still 36 feet total. What dimensions maximize the area, and how would a fourth grader find them without algebra?
Read the worked answer
With two widths (the two ends) and one length, 2w + L = 36, so L = 36 − 2w and area = w(36 − 2w).
Table of possibilities:
| Width (each end) | Length | Area (ft²) |
|---|---|---|
| 3 | 30 | 90 |
| 6 | 24 | 144 |
| 9 | 18 | 162 |
| 12 | 12 | 144 |
| 15 | 6 | 90 |
The maximum is width 9 ft, length 18 ft, area 162 ft² - more than double the 81 ft² of the four-sided square, because the wall supplies one side free.
The pattern is symmetric: 144 appears twice (6 × 24 and 12 × 12), so the peak must sit between them at 9. A fourth grader can find it by building the table systematically, noticing the symmetry, and testing neighbors 8 and 10 to confirm neither does better. Worth highlighting for the class: 18 is twice 9, and 36 is twice 18 - the same "spread the fencing evenly" reasoning that made the square win in the four-sided version.