Friday night we threw an house warming party and invited quite a number of fellow students. To fit everybody around the table we had to enlarge it, pulling out two sort of shelves from the short sides, but unfortunately our tablecloth was too small.
Being a bunch of smart engineering students we started to discuss how to lay out the tablecloth on the table, whether an optimum existed and how to find it. Luckily enough beer and wine let us rapidly change the subject, but did not kill the curiosity.
Here is some (sorta) math jargon, finally:
- you have a table of known dimensions $a$ and $b$
- you have a tablecloth of known dimensions $c$ and $d$
- $a>c>d>b$ holds$^*$
Thickness is not a problem, i.e. the tablecloth might hang how much you want and the table sides must not necessarily be covered. This is a problem on the plane.
Your goal is to maximize the covered surface of the table, your knobs are three real numbers, the tablecloth position and its angle with respect to the table.
I am afraid that the optimum depends on how the inequality holds, i.e. the solution changes dramatically if the tablecloth short side is, say, 100 times the table short side. If this is the case please stick to "reasonable" values, i.e. the difference between the various dimensions is within the $10\%$ range. And I also believe the solution is symmetrical, i.e. the tablecloth is centered on the table, but I am not entirely sure of it.
Folding the tablecloth is not permitted, mainly because the problem gets heavily dependent on the ratio $\frac{d}{b}$.
The question is:
How can I prove there is an optimum and find it?
$^*$this means that the tablecloth short side is longer than the table short side, and the table long side is longer than the tablecloth long side. Plus both the table and the tablecloth are rectangular.
If you place the tablecloth naively, i.e. aligned with the table, you cover an area $A(0) = bc$.
If you rotate the tablecloth by an angle $\theta$ without any corner crossing any edge, the area increases to $A(\theta)=bc\sec\theta$.
The best you can do with this strategy is when two opposite corners of the tablecloth are on the long edges of the table*. In this case, letting $p=\sqrt{c^2+d^2}$, the angle is $$\theta^*=\sin^{-1}\left(\frac dp\right)-\sin^{-1}\left(\frac bp\right),$$ and the area is $$A(\theta^*) = \frac{bcp^2}{bd+\sqrt{p^2-d^2}\sqrt{p^2-b^2}} = \frac{bc(c^2+d^2)}{bd+c\sqrt{c^2+d^2-b^2}}.$$ I conjecture that this is the optimal solution. However, if all the dimensions are within $10\%$ of each other, this buys you less than $0.5\%$ more area than the naive solution (when $b=0.9c$, $d=c$).
*Assuming the table is long enough. I forgot to consider the case when the diagonal of the tablecloth is longer than that of the table, but the analysis of that should be symmetrical.