Fencing a Garden with 24 Meters: Which Design Holds the Most?
A third-grade class is designing rectangular garden beds in the school courtyard. Each group receives exactly 24 meters of fencing and must use all of it for the perimeter. Group A proposes a square 6 meters by 6 meters. Group B proposes 8 meters by 4 meters. Group C proposes 11 meters by 1 meter. The beds must also be at least 3 meters wide so students can reach the middle without stepping on the plants. Every group must prove its perimeter using 2 × (length + width) and its area using length × width. Midway through, a student asks whether a group could design a bed with an area of exactly 40 square meters using the same 24 meters of fencing, and another asks whether the side lengths have to be whole numbers at all. The teacher wants both questions settled with reasoning rather than by guessing or by trusting whoever answers first.
1. Check each of the three designs for perimeter, area, and the 3-meter width rule, then state which design the class should build.
Read the worked answer
Group A: perimeter 2 × (6 + 6) = 24 m; area 6 × 6 = 36 square meters; width 6 m passes the rule. Group B: perimeter 2 × (8 + 4) = 24 m; area 8 × 4 = 32 square meters; width 4 m passes. Group C: perimeter 2 × (11 + 1) = 24 m; area 11 × 1 = 11 square meters; width 1 m fails the 3 m rule.
Group C is disqualified on width, and Group A's 36 square meters beats Group B's 32. The class should build the 6-by-6 square: same fencing, most growing space.
2. Can a rectangle with a 24-meter perimeter have an area of exactly 40 square meters? Support the answer by listing the whole-number side pairs, and say whether allowing fractional sides changes the conclusion.
Read the worked answer
For a perimeter of 24, length + width = 12. List the whole-number pairs and their areas:
1 and 11 → 11; 2 and 10 → 20; 3 and 9 → 27; 4 and 8 → 32; 5 and 7 → 35; 6 and 6 → 36.
No pair gives 40, and the areas grow as the sides get closer together, peaking at the square with 36 square meters. Since 36 is already the largest value in the pattern and 36 < 40, area 40 is impossible.
Fractional sides do not rescue the claim. A 5.5 m by 6.5 m bed still has perimeter 24, but its area is 35.75 square meters, which is smaller than 36 and still far from 40.
3. Group D wants a bed 9 meters long. Find the width it must have, check it against the width rule, and explain what happens to the area when a design becomes long and thin.
Read the worked answer
Perimeter 24 means 2 × (9 + width) = 24, so 9 + width = 12 and width = 3 m.
Width check: 3 m exactly meets the minimum, so this design is allowed, but only just.
Area: 9 × 3 = 27 square meters, compared with 36 for the square. The fencing is identical; the long, thin bed simply wastes more of the perimeter on two long sides that enclose little space. Stretching a rectangle while holding the perimeter fixed trades away area.