# Height and width of x rectangles to fit in area

I have an area of $A = 320 \times 480$. I then have a number of rectangles of max size $128 \times 152$ which need to fit into the area. There can either be $1$ rectangle or $100$ rectangles. How would calculate the size that the rectangles need to be to fit into the given area?

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Can you see how $100$ $32\times48$ rectangles would fit? – Gerry Myerson Feb 11 '13 at 8:51
What would that equation look like if I may ask? – Cam Feb 11 '13 at 9:03
What equation? --- I'm talking geometry, not algebra. Take your big rectangle $A$, slice it in $10$ equal slices vertically, and in $10$ equal slices horizontally, and you get $100$ rectangles, each one-tenth as long and one-tenth as wide as $A$. – Gerry Myerson Feb 11 '13 at 12:04

Given a number $n$ and a big rectangle of size $a\times b$, you want to find a rectangle size $x\times y$ such that $n$ such rectangles fit exactly into the big rectangle. A simple method to achieve this, is to find a factorization $n=m\cdot k$ of $n$ and let $x=\frac am$, $y=\frac bk$. This way it is possible to lay the small rectangles in a regular $m$ by $k$ pattern and thus fill the big rectangle.
For example, if $a=320$, $b=480$ as in your example and $n=100$, one might choose the factorization $n=10\cdot 10$ and thus arrange $10$ by $10$ rectangles of size $32\times 48$. Or we might factor $n=5\cdot20$ and arrange $5$ by $20$ rectangles of size $64\times 24$. Each factorization $100=1\cdot 100=2\cdot 50=4\cdot 25=5\cdot 20=10\cdot 10=20\cdot5=25\cdot 4=50\cdot 2=100\cdot 1$ gives rise to a different solution.
An interesting question arises: If we allow the small rectangles to be rotated by $90^\circ$ and thus use "mixed" patterns, could solutions with other sizes than those found by the method above come up? The somewhat surprising (partial) answer is: One of the fractions $\frac ax$, $\frac bx$ must be an integer, and one of the fractions $\frac ay$, $\frac by$ must be an integer (this can be shown using a clever integration trick). In special cases (e.g. if the given rectangle is a square, $a=b$), this already shows that $x\times y$ is among the solutions found with the factorization method described above.
You said, "There can either be $1$ rectangle or $100$ rectangles," so where do these $41$ rectangles come from? Anyway, one way to do it would be to use rectangles that are $320\times(480/41)$. – Gerry Myerson Feb 11 '13 at 12:08