I initially assumed that this problem could be solved with formulas, but now I'm not so sure.
The simply version can be solved easily: Consider a workshop that must complete two jobs. Job A takes 10 full days worth of work, and Job B takes 5 full days worth of work. The restrictions are that:
- the portions of resources devoted to each job must be the same on each day of work; and
- both jobs must finish at the same time.
What portion of each day should be spent on each job, and after how many days will the jobs be finished?
Let:
- a = # full days of work for Job A
- b = # full days of work for Job B
Then:
- Daily portion for Job A = $$\frac{a}{a + b} = \frac{10}{10 + 5} = 0.\overline{666}$$
- Daily portion for Job B = $$\frac{b}{a + b} = \frac{5}{10 + 5} = 0.\overline{333}$$
- (And of course:) $$0.\overline{333} = 1 - 0.\overline{666}$$
- Number of days to finish both jobs = $$a + b = 10 + 5 = 15$$
The more complex version... Suppose the workshop has higher proficiency for Job A during the first 5 days, so any resources devoted to Job A yield double the results. But the portions for each job must still be the same on each day during the whole period of work.
I tried to solve this in a spreadsheet using formulas, but could not. I was able to use formulas to determine how much work would be accomplished cumulatively on all the days of work given Job A/B portions that I specify. This allowed me trial-and-error discover portion values that result in both jobs finishing approximately early in the 13th day.
But could this be solved directly and exactly using a formula?
For these kind of problems, dimensional analysis can help clarify things. Here's one way to frame the problem with some formality (perhaps a bit excessive) but hopefully more interpretability.
Suppose job $A$ requires $a$ units of output and job $B$ requires $b$ units of output. Suppose the firm's overall productivity is $\Pi$ units/day. Then the information that is given can be summarized as
$$10 \text{ days}={a \text{ units} \over \Pi {\text{units}\over \text{day}}}\iff 10=a/\Pi\\ 5 \text{ days}={b \text{ units} \over \Pi {\text{units}\over \text{day}}}\iff 5=b/\Pi.\\$$
Your aim is to find a portion, $k\in (0,1)$, of the firm's productivity to allocate to job $A$ (with the remaining portion $1-k$ allocated to job $B$) so that the completion time for both jobs is the same.
Now if any productivity allocated toward job $A$ is magically doubled within the first 5 days, it will still take at least $5$ days to complete job $A$ since $${a \text{ units} \over 2k\Pi {\text{units}\over \text{day}}}\geq {a \text{ units} \over 2\Pi {\text{units}\over \text{day}}}=5 \text{ days}.$$
After the first $5$ days, there is no magic productivity boost, and the amount of work remaining on job $A$ is $$a\text{ units}-2k\Pi{\text{units}\over \text{day}}\times 5 \text{days}.$$ Thus, your problem is to solve for $k\in (0,1)$ so that completion times are equal:
$$\begin{align}\text{time for job }A&=\text{time for job }B\\ 5 \text{ days}+{a\text{ units}-2k\Pi{\text{units}\over \text{day}}\times 5 \text{days} \over k\Pi {\text{units}\over \text{day}} }&={b \text{ units} \over (1-k)\Pi {\text{units}\over \text{day}}} \\ {10 \over k}-5&={5 \over 1-k}\\ k^2-4k+2&=0\\ k&=2-\sqrt{2} \end{align}$$
Note the other solution of the quadratic, $k=2+\sqrt{2}$, is invalid (it exceeds one).
Thus, the time spent to complete both jobs is ${5 \over 1-k}={5 \over \sqrt{2}-1}\approx 12.07,$
or early in the $13$th day, which agrees with your answer.