Number of lines |
New rectangles |

1 | 0 |

2 | 0 |

3 | 0 |

4 | 1 |

5 | 1 |

6 | 2 |

7 | 2 |

8 | 3 |

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

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.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.