Tukutuku Panel: Fitting a Growing Pattern Inside a Fixed Frame
At a wharekura, ākonga are weaving a tukutuku panel for the new whare. The pattern is built row by row from the bottom: the bottom row has 7 stitches, and each row above it has 4 more stitches than the row below. The stitches of a row sit in one horizontal line, and each row is 6.0 cm high, including the gap to the row above.
The panel frame is 0.95 m wide and 1.20 m tall. Each stitch, including the gap beside it, takes up 1.6 cm of the row's width. A finished row must fit inside the 0.95 m width with no stitch cut off, and every row must fit inside the 1.20 m height.
The kaiako wants the pattern to fill the frame as fully as possible in both directions and needs to know the total number of stitches to prepare before weaving starts, because each stitch is a separate cross-stitch that must be counted out and cut in advance.
1. Write a rule for the number of stitches in row n, counting the bottom row as row 1. Explain what each number in your rule represents and why the pattern is linear.
Read the worked answer
Rule: S = 7 + 4(n − 1), which simplifies to S = 4n + 3
Where each number comes from
- 7 is row 1 - the starting value.
- 4 is the common difference: every row adds 4 stitches.
- (n − 1) counts how many steps up from row 1 you have moved. Row 1 has had no steps, so it must give 7, not 11 - this is the most common slip.
Why it is linear: each row is the row below plus a fixed 4 stitches, so the number of stitches changes by the same amount every time. A constant difference means a straight-line (linear) relationship with gradient 4 and intercept 3.
Check by listing: 7, 11, 15, 19, 23 …
- n = 1: 4(1) + 3 = 7 ✓
- n = 4: 4(4) + 3 = 19 ✓ matches the list
Because the pattern grows, the widest row is always the top row - that row, not the bottom one, determines whether the panel fits.
2. Find the greatest number of rows the pattern can have inside the 0.95 m × 1.20 m frame. Show both the width and the height check.
Read the worked answer
Width check
- Frame width in centimetres = 0.95 m = 95 cm
- Stitches that fit = 95 ÷ 1.6 = 59.375, so the widest row can have at most 59 stitches (60 would be 96 cm and would be cut off).
Set 4n + 3 ≤ 59
- 4n ≤ 56
- n ≤ 14
So the width allows 14 rows, and row 14 would have 4(14) + 3 = 59 stitches - exactly at the limit.
Height check
- Frame height in centimetres = 1.20 m = 120 cm
- Rows that fit in height = 120 ÷ 6 = 20 rows
Conclusion: height allows 20 rows, but width allows only 14. The width is the limiting constraint, so the greatest number of rows is 14.
Check row 15: 4(15) + 3 = 63 stitches, and 63 × 1.6 = 100.8 cm - 5.8 cm too wide. Confirmed: 14 is the maximum.
Note that with 14 rows the panel is 14 × 6 = 84 cm tall, leaving 36 cm of unused height. The frame is not being filled vertically, which the kaiako may want to address by widening the pattern rather than adding rows.
3. Calculate the total number of stitches that must be prepared for the maximum-height design of 14 rows. Show a method that does not require adding every row one at a time.
Read the worked answer
Rows 1 to 14: 7, 11, 15, …, 59 (14 terms, common difference 4)
Method - pair the ends
- First + last = 7 + 59 = 66
- Second + second-to-last = 11 + 55 = 66
- Every pair has the same total, and 14 terms make 7 pairs
- Total = 7 × 66 = 462 stitches
Method - average × number of rows
- Average row = (7 + 59) ÷ 2 = 33
- Total = 33 × 14 = 462 stitches ✓
Cross-check using the rule S = 4n + 3
- Sum of 4n for n = 1 to 14 is 4 × (1 + 2 + … + 14) = 4 × 105 = 420
- Sum of the 3s is 3 × 14 = 42
- Total = 420 + 42 = 462 ✓
Answer: 462 stitches.
This figure matters practically: 462 separate cross-stitches must be cut and counted before weaving, and running short partway up the panel would leave an unfinished row at the top where the pattern is widest and most visible. Preparing 470 would allow for a few damaged stitches, but the panel itself must stay at 14 rows and 462 stitches to fit the frame.