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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?.

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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.

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  • $\begingroup$ Actually this was what I intended to avoid guessing as well I'm not very good at it and the request in my question was to know if there was perhaps a more systematically way to address these kinds of problems. I've tried to use a recommendation which used systems of equations. But I don't know if this strategy can work in all cases, hence looking for a better approach or to proof if what I was doing was right. $\endgroup$ – Chris Steinbeck Bell Feb 13 at 18:14
  • $\begingroup$ I suppose that you use the $11$ pounds from the first trial as a weight plus the $7$ weight again in one pan and fill out the other with the remaining lactose from the flask until balances to $18$ this confirms to do in two trials. But I don't understand the second portion of your passage. Splitting evenly, do you meant obtaining $50$ pounds from the first trial but what to do with that after?. The same applies with the $52$ which comes from $104$. $\endgroup$ – Chris Steinbeck Bell Feb 13 at 18:20
  • $\begingroup$ I usually try to find some solution to start, not necessarily the best. Once we have the two step solution we can ignore anything but one step solutions. There are no so many of those. My suggestion for splitting the $100$ was just my way of seeing what the information that we start with $100$ gives us. It turned out not to help here, but it might have. To be a good puzzle, there should be a clever and nonobvious solution, like splitting the $100$. $\endgroup$ – Ross Millikan Feb 13 at 18:23
  • $\begingroup$ Okay. I understood that. But does splitting the $100$ helps in this case?. From your answer it seems not. But in the previous comment it gave me the impression that $50$ or $52$ could be used to get a solution. Because I'm trying all sorts of hand calculations and I can't find a way to use the $50$ to obtain the desired $18$ neither the $52$. This part is where I'm stuck at. $\endgroup$ – Chris Steinbeck Bell Feb 13 at 18:28
  • $\begingroup$ There is also an unattended question which is. Does a situation involving finding the least of something does it always lead to use of derivatives?. It seems doesn't apply in this case. Sorry if it sounds redundant but I don't want to get confused and act automatically each time I see some tag word in a problem. $\endgroup$ – Chris Steinbeck Bell Feb 13 at 18:34

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