For one of my exams I am allowed a one-sided equation sheet on a regular piece of printer paper. This got me thinking.
If I write in rows, that is, I start at the top right and write until I get to the end, and then continue doing that, is it more beneficial to write portrait or landscape?
Can one fit more information by flipping the page a certain way, or is it all the same?
If one way does let me write more, is there mathematics behind this?
I know - this is the epitome of procrastination.
Nice question, here's one possible interpretation.
Let $W(l)$ be the amount of words/equations/whatever you can fit on a row of length $l$ (in whatever units suit you). $W$ depends on how wide you write your letters. Let $H( r)$ be the number of rows of text you can fit in a column of length $r$. $H$ depends on how tall you write your letters.
Consider a piece of paper that has dimensions $a\times b$ (with $a,b$ in the same units as $l$ and $r$). Suppose WLOG that $a \geq b$.
When orienting the paper in landscape, the amount you can write is $W(a)H(b)$ because you have $H(b)$ rows each with an amount $W(a)$ of writing. In portrait, the amount you can write is $W(b)H(a)$. So our decision rule is: choose landscape if $W(a)H(b) > W(b)H(a)$.
Let's further assume that $W$ and $H$ are linear in terms of their respective arguments, i.e.
\begin{align} W(l) &= \frac{l}{\omega}\\\\ H( r) &= \frac{r}{\eta}\\ \end{align}
where $\omega$ is the (average) width of one letter you write and $\eta$ is the (average) height of one letter you write.
Using the inequality from before, we obtain \begin{align} W(a)H(b) &> W(b)H(a)\\ \frac{ab}{\omega\eta} &> \frac{ba}{\omega\eta}\\ 1 &> 1 \end{align}
So using our decision rule, we see that we should use portrait since the inequality does not hold. However, we also see that the amount we can write is the same in both cases, so it doesn't matter.
This is an interesting exercise in modeling and is instructive: by making some assumptions about your model you can derive certain results, which only hold as long as the model itself and your assumptions are valid. I'm sure someone can come along and find a better model for the amount you can write which may produce different results. For example, one assumption that I made was that the "amount" you can write in a row or column is a continuous quantity, when in reality it is probably a positive integer measured in words.
One standard way to extend the answer by @Alex is to use integer math instead of continuous math, e.g. the floor function $\lfloor \cdot \rfloor$ like this:
\begin{align} W(l) &= \lfloor \frac{l}{\omega} \rfloor \\\\ H( r) &= \lfloor \frac{r}{\eta} \rfloor \\ \end{align}
This becomes the problem of packing small $\omega \times \eta$ rectangles (all in the same orientation) into a large $l \times r$ rectangle. One orientation might pack more than the other.
Alternatively, we could model that some math symbols are taller than others, like ${1 \over \sqrt{x^2 + y^2}}$. Suppose we want the height of each row to be the height of the tallest symbol (like how this post is being formatted). This would generally (but not always) argue for shorter rows (i.e. writing in portrait) because each maximization is over a smaller set. Under simple probabilistic input models you can prove that portrait is probabilistically better. However for a specific input you'd just have to try it out both ways to see which is better.
Alternatively, we could model that a word will not be broken up across successive lines (i.e. hyphe-nation like this is not allowed). This would generally (but not always) argue for longer rows (i.e. landscape) because you have fewer potential wastages at the end of lines. Again, you can prove various probabilistic statements but for a specific input you'd just have to try it out both ways to see which is better.
