### by Howard Williams

#### Published Sunday August 22 2021 (link)

Without changing their size, Judith sews together one-foot squares of different colours that her mother has knitted, to make rectangular throws. These are usually all of the same dimensions using fewer than a hundred squares. She has observed that it takes her mother 20 per cent longer to knit each square than it takes her to sew two single squares together.

As a one-off she has completed a square throw whose sides have the same number of squares as the longer side of her usual rectangular throws. The average time it took per square foot, both knitting and sewing, to complete the square throw was 2 per cent longer than that of the rectangular throws.

What are the dimensions in feet of the rectangular throws?

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Here is a manual solution. Let the throws have the dimensions $$m$$ and $$n$$, where $$m$$ is the larger side.  The area is them $$mn$$ and the number of sewn edges is $$(2mn\;-\;(m + n))$$. Using a time unit that makes the time to sew an edge 5 units and that to knit a square 6 units, the time to sew a throw is $$6mn + 5(2mn\;-\;(m + n)) = 16mn\;-\;5(m + n)$$. The square throw takes 2% longer per square foot to make than a rectangular throw, hence:

$\left(\frac{102}{100}\right)\frac{16mn\;-\;5(m + n)}{mn} = \frac{16m^2\;-\; 10m}{m^2}$

This can be simplified to show that:

$m=\frac{245}{255/n\;-\;16}$and $16\;+\frac{5\times7^2}{m}\;=\;\frac{3\times5\times17}{n}$

The first equation shows that $$m<20$$ when $$n<10$$ (which we know) so we can see from the second equation that $$(m, n)$$ is one of $$(5, 3)$$, $$(7, 3)$$ or $$(7, 5)$$, of which only $$(7, 5)$$ provides a solution. Hence the throws are 7 by 5 feet.

2. Surely it is the smaller dimension that is less than 10.

e.g. A throw with dimensions 3 × 33 has area less than 100.