Fencing a garden against the school wall with a fixed 36 m roll
The Grade 6 class is laying out a rectangular vegetable garden against the straight back wall of the school. The wall forms one side and needs no fencing; the other three sides will be fenced with the 36 m of fencing the head teacher has approved. The whole roll must be used, with no cutting except at the corners. The class must decide the two dimensions. A pupil draws a square 12 m by 12 m and argues that 'a square always gives the biggest area'. The garden will also be edged along the wall side with square paving stones, each 50 cm by 50 cm, sold only in packs of 8. The class measures with a tape marked in metres and centimetres; one group records the long side as 18 m when the tape actually reads 17 m 85 cm.
1. Find the dimensions that give the greatest area, and explain why the 12 m by 12 m square is not the best answer.
Read the worked answer
Let the two sides coming out from the wall each be w metres. The side along the wall is then 36 − 2w, because the two widths plus the long side use the whole roll.
| w (m) | Long side 36 − 2w (m) | Area w × (36 − 2w) |
|---|---|---|
| 6 | 24 | 144 m² |
| 8 | 20 | 160 m² |
| 9 | 18 | 162 m² |
| 10 | 16 | 160 m² |
| 12 | 12 | 144 m² |
| 14 | 8 | 112 m² |
The greatest area is 9 m by 18 m = 162 m², with 9 + 18 + 9 = 36 m of fencing.
The square uses 12 + 12 + 12 = 36 m too, but gives only 144 m² - 18 m² less. The square rule only applies when all four sides are fenced. Here the wall is a free side, so the fencing is shared over three sides and the best rectangle is twice as long as it is wide. The areas rise to 162 m² and fall away again either side of w = 9, which is why 9 m is the peak.
2. How many paving stones are needed along the 18 m side, and how many packs must be bought? What if the class had chosen the 20 m side instead?
Read the worked answer
Units first: the stones are measured in centimetres, the garden in metres, so convert. 18 m = 1,800 cm.
Stones along 18 m: 1,800 ÷ 50 = 36 stones exactly, so no cutting is needed.
Packs: 36 ÷ 8 = 4.5. Packs are sold only in 8s, and 4 packs (32 stones) would leave a 4-stone gap, so the class must buy 5 packs = 40 stones, with 4 spare.
If the 20 m side were chosen: 20 m = 2,000 cm; 2,000 ÷ 50 = 40 stones; 40 ÷ 8 = 5 packs exactly, no waste. This is a reminder that a cheaper-looking layout can waste material, but it does not change the answer to the area question: the 8 m by 20 m garden has area 160 m², which is still 2 m² smaller than 9 m by 18 m. The class must decide which matters more - maximum growing area or no leftover stones.
3. One group recorded the long side as 18 m when the tape read 17 m 85 cm. How much does this error change the calculated area, and does it matter?
Read the worked answer
The width is 9 m in both cases; only the long side is misread.
- Recorded area: 9 × 18 = 162.00 m²
- True area: 9 × 17.85 = 160.65 m²
- Difference: 162.00 − 160.65 = 1.35 m²
The short way to see it: the error is 18 m − 17.85 m = 0.15 m, and each 0.15 m of extra length adds 9 × 0.15 = 1.35 m².
Does it matter? As a fraction of the area it is 1.35 ÷ 160.65 ≈ 0.8%, under 1%, so the choice of 9 m by 18 m as the best shape is not affected - 162 m² and 160.65 m² are both well clear of the 160 m² of the 8 m by 20 m design. But it does matter for buying seed, fertiliser or topsoil sold by area, and it matters more if the same 15 cm slip happens on several sides. The class should re-measure and record in centimetres, not round to the nearest metre.