How Many Rectangles?
By drawing 5 horizontal and four vertical lines, one can form 12 rectangles. What is the greatest number of rectangles that can be formed by drawing 15 lines?
Problem
By drawing $9$ lines, $5$ horizontal and $4$ vertical, one can form $12$ small rectangles, as shown on the right. What is the greatest possible number of small rectangles that one can form by drawing $15$ lines, either horizontal or vertical?
If you liked this problem, here is an NRICH task that challenges you to use similar mathematical ideas.
Student Solutions
Answer: 42
The sequence of patterns
| Number of lines | New rectangles |
| 1 | 0 |
| 2 | 0 |
| 3 | 0 |
| 4 | 1 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
| 8 | 3 |
You add more rectangles horizontally, then vertically, so the pattern continues.
9th line adds 3 rectangles (already seen that 9 lines make12 rectangles)
10th, 11th lines +4 each total 12 + 8 = 20
12th, 13th lines +5 each total 20 + 10 = 30
14th, 15th lines +6 each total 30 + 12 = 42
An expression for the number of rectangles
Suppose there are $a$ horizontal and $b$ vertical lines. The grid of rectangles formed is then $a-1$ rectangles high, and $b-1$ rectangles wide. This means there are $(a-1)(b-1)$ rectangles.
If there are a total of $15$ lines, the aim is to make $(a-1)(b-1)$ as large as possible with $a+b=15$.
This can be done by considering the different combinations that add to make $15$:
| $a$ | $b$ | $(a-1)(b-1)$ |
|---|---|---|
| $1$ | $14$ | $0 \times 13 = 0$ |
| $2$ | $13$ | $1 \times 12 = 12$ |
| $3$ | $12$ | $2 \times 11 = 22$ |
| $4$ | $11$ | $3 \times 10 = 30$ |
| $5$ | $10$ | $4 \times 9 = 36$ |
| $6$ | $9$ | $5 \times 8 = 40$ |
| $7$ | $8$ | $6 \times 7 = 42$ |
Therefore, the largest number is $42$ rectangles, formed by having seven lines in one direction and eight in the other.
Alternatively, you can use completing the square to maximise the quantity:
Since $a+b=15$, $(a-1)(b-1) = (a-1)(14-a) = -a^2+15a-14$. Then, by completing the square, this is $-\left(a-\frac{15}{2}\right)^2 + \frac{169}{4}$.
This is maximised when the square is minimised, which occurs when $a=7$ or $a=8$ (since $a$ must be an integer). This gives $6 \times 7 = 42$ rectangles.