I've been going in circles with this kind of problem, hence need some help into trying to find if this kind of situation can be solved without the use of derivatives.
The problem is as follows:
In a medical laboratory in Hsinchu a technician has been tasked to weigh $\textrm{100 lb}$ of pure grade lactose from a flask to be later collected and used in several blood analysis for patients. To do this he has to use a two-pan scale, however in the lab bench there are only three weights one of $\textrm{4 lb}$, the second of $\textrm{7 lb}$ and a third one of $\textrm{13 lb}$. What is the minimum number of trials does he has to do in order to fill a bottle of $\textrm{18 lbs}$ of lactose?
The alternatives given in my book were:
$\begin{array}{ll} 1.&1\,\textrm{trial}\\ 2.&3\,\textrm{trial}\\ 3.&4\,\textrm{trial}\\ 4.&5\,\textrm{trial}\\ 5.&2\,\textrm{trial}\\ \end{array}$
Upon revising my notes I found that supposedly the recommendation into trying to solve is to build up a system of $2 \times 2$. This meant that all the weights must be used altogether. First I thought that from the start I wouldn't worry about the first quantity mentioned or the $\textrm{100 lb}$, this information was given to say just the upper boundary of what the technician can use but it seems not needed for calculations.
Therefore:
$\begin{array}{l} x+y=4+7+13=24\\ x-y=18\\ \end{array}$
From this is established that:
$$2x= 42$$
$$\textrm{x= 21 lb}$$
$$\textrm{y= 3 lb}$$
However this arises the problem that there are not weights of $\textrm{21 lb}$ or $\textrm{3 lb}$. So it looks that it cannot be obtained that specific weight by one trial.
Therefore I asked. By going on a second trial then I would use two weighs in the first trial the $\textrm{13 lb}$ and the $\textrm{7 lb}$ to obtain $\textrm{20 lb}$, which would be "converted" as if it were a new weight to be used in the second attempt. I decided not to use the other quantity of $\textrm{3 lb}$ as there isn't a lower weight to attain that mass.
So by the second trial would be: (again using the rationale of using all the weights)
$\begin{array}{l} x+y = 4+7+13+20=44\\ x-y=18\\ \end{array}$
From this is obtained that:
$$2x= 62$$
$$x= 31$$
$$y=31-18=13$$
Now with this result there is a weight of $\textrm{13 lb}$. Therefore it can be obtained on a second trial by putting that mass along with a small pile of the lactose from the flask so that:
In the right side of the pan using the $4$ and $7$ weights:
$20+4+7=31$
In the left side of the pan using the $\textrm{13 lb}$ weight:
$x+13$
So that when the scale is used:
$\textrm{31=x+13}$
therefore $\textrm{x=18 lb}$
and this is how I got to the $\textrm{18 lb}$ but in the second trial. From this I concluded that the reasonable answer would be the alternative $5$ in the set of answers in other words, the technician must use at least $\textrm{2 trials}$ to fill the bottle with lactose. But I'm not very sure if my answer is the right one or if this method is the faster one.
Therefore I would like somebody could help me with this matter and assist me on if there is an alternate method rather than just plugging in numbers randomly or guessing. So overall can somebody help me?. Needless to say that each time I read the word least I immediately think that derivatives may be involved, but is this the case?.
I don't think there is a way besides clever guessing for problems like this. Here we note that $7+7+4=18$ so he can weigh it in two trials-weigh $7+4$ once and $7$ the second time. I would take the $100$ to be the amount he starts with, so you could weigh $50$ in one weighing by splitting it evenly, or $52$ by putting the $4$ on one pan and splitting the lactose to make it balance, but I couldn't find a way to get $18$ in one weighing.