Relationships Between Area and Perimeter
Warm-up
Two rectangles: 2x10 (perimeter 24, area 20) and 4x6 (perimeter 20, area 24). Neither has both the larger area AND the smaller perimeter. The shape that minimized the perimeter also maximized the area. Is this always true? Today we investigate.
Explore
Systematic investigation: each group fixes one measurement (perimeter OR area) and varies the rectangle dimensions. Record all results in a table. Graph: one variable on each axis. Share findings: which shape gives the most area for a fixed perimeter? Which gives the least perimeter for a fixed area? (Both: the square.)
Formalize
Garden design challenge: you have 60 m of fencing. Design a rectangular garden that maximizes growing space. What dimensions? What is the maximum area? (15x15=225 m squared. Compare to 10x20=200, 5x25=125.) The square wins. If the garden must be rectangular (not square), which rectangle comes closest?
Relationships Between Area and Perimeter
Economic connection: building a wall costs money. If each metre of wall costs 8,000 in walls. vs. 4x25 = 58 m = $11,600.) Real design has real monetary implications.
Practice
Students complete one fixed-perimeter and one fixed-area investigation, both recorded in tables and graphed. Solve 2 real garden-design problems. Exit ticket: for perimeter 32 cm, what rectangle dimensions give the greatest area?
Exit ticket
Students complete one fixed-perimeter and one fixed-area investigation, both recorded in tables and graphed. Solve 2 real garden-design problems. Exit ticket: for perimeter 32 cm, what rectangle dimensions give the greatest area?
Step 1: The farmer's constraint: 24 m of fence, rectangular pen, whole-metre sides. List EVERY rectangle: sides must sum to 12 per pair → 1×11, 2×10, 3×9, 4×8, 5×7, 6×6.
Step 2: Compute all areas: 11, 20, 27, 32, 35, 36 m².
Step 3: Graph area against the first side (bar chart, sides 1–6): the bars climb and crest at the square. Not a line — a curve that RISES FAST then flattens near the top (35 to 36 is a whisper; 11 to 20 was a shout).
Step 4: State the discovered theorem: for a fixed perimeter, the square maximizes rectangular area — and near-squares are nearly as good. Skinny pens hemorrhage area.
Step 5: The half-fence twist (a real classic): the pen backs onto a barn — fence only 3 sides with the same 24 m. Now sides w + w + L = 24. Try: 4×16 = 64(!), 5×14 = 70, 6×12 = 72, 7×10 = 70. Best is 6×12 — NOT a square, and the area beats every 4-sided pen. Changing constraints changes the champion; there is no formula-reflex that replaces exploring.
Step 1: The gardener's constraint flipped: the bed must COVER exactly 36 m²; fencing is bought after. Rectangles with area 36: 1×36, 2×18, 3×12, 4×9, 6×6.
Step 2: Perimeters: 74, 40, 30, 26, 24 m.
Step 3: Read the table's story: same soil, from 24 m of fence to 74 m — the 1×36 sliver costs THREE TIMES the fencing of the 6×6 square. And it can get worse: allow half-metres — 0.5 × 72 needs 145 m! With fixed area, perimeter has NO ceiling; it only has a floor (the square, again).
Step 4: Twin theorems, side by side on the anchor chart: fixed perimeter → square maxes area; fixed area → square minimizes perimeter. One shape, champion of both contests — compactness is its superpower.
Step 5: Nature's vote: why are soap bubbles round, cells roundish, igloos domed? Minimizing boundary for the space enclosed saves material, heat, and effort. The class's fence tables and a soap bubble are making the same decision.
The claim to test: student adds a bump-out to a rectangle ("I made the perimeter bigger, so the area must be bigger too").
Step 1: Build the counterexample together: start with a 6×4 rectangle (P = 20, A = 24). Now cut a 2×2 notch INTO one side. Walk the new boundary: the notch ADDS two edges of 2 (going in and coming out) → P = 24. But the notch REMOVED area → A = 20.
Step 2: Sit with it: perimeter UP, area DOWN, same object. The claim is dead — one counterexample suffices (a lovely piece of mathematical logic in itself).
Step 3: Explore the full 2×2 grid of possibilities and fill it with examples: P↑A↑ (enlarge the whole thing), P↑A↓ (the notch), P↓A↓ (shrink it), P↓A↑ — possible?? Squash a 1×11 (P=24, A=11) into a 4×5 (P=18, A=20): perimeter down, area UP ✓.
Step 4: The moral, earned the honest way: perimeter and area are independent quantities — neither drags the other. Any "more of one means more of the other" instinct dies today, by counterexample, in all four directions.
Exit: produce your own P↓A↑ pair different from the class's.