Designing a Pen for the Class Rabbit
The class is building a rectangular pen for a rabbit against a straight wall. They have 20 metres of fencing to use on three sides; the wall itself forms the fourth side and needs no fencing. Students sketch several rectangles and measure the area each one would enclose. One student draws a pen 2 m deep and 16 m long along the wall. Another draws a pen 5 m deep and 10 m long. A third draws a pen 6 m deep and 8 m long. The class must use all 20 m of fencing and wants the rabbit to have the largest possible area.
1. Check that each of the three sketched pens uses exactly 20 m of fencing, then calculate the area of each. Which sketch gives the largest area so far?
Read the worked answer
For a pen against a wall, fencing covers two depths and one length: fencing = 2 × depth + length.
- 2 m deep, 16 m long: 2 × 2 + 16 = 4 + 16 = 20 m ✓. Area = 2 × 16 = 32 m².
- 5 m deep, 10 m long: 2 × 5 + 10 = 10 + 10 = 20 m ✓. Area = 5 × 10 = 50 m².
- 6 m deep, 8 m long: 2 × 6 + 8 = 12 + 8 = 20 m ✓. Area = 6 × 8 = 48 m².
All three use exactly 20 m. The 5 m by 10 m pen has the largest area so far, 50 m².
2. Try two more whole-metre depths between 4 m and 7 m, calculate their areas, and decide which depth gives the greatest area overall. Explain how the areas change as the depth grows.
Read the worked answer
Use fencing = 2 × depth + length = 20, so length = 20 − 2 × depth.
- Depth 4 m: length = 20 − 8 = 12 m. Area = 4 × 12 = 48 m².
- Depth 7 m: length = 20 − 14 = 6 m. Area = 7 × 6 = 42 m².
Collect the results:
- 2 m → 32 m²
- 4 m → 48 m²
- 5 m → 50 m²
- 6 m → 48 m²
- 7 m → 42 m²
The area rises as depth grows from 2 m to 5 m, reaches a peak of 50 m² at 5 m, then falls again. So a depth of 5 m with a length of 10 m gives the greatest area, 50 m².
Reasoning: as depth increases, length shrinks by twice as much, so beyond the middle the loss in length outweighs the gain in depth.