Htykuut

In a certain video game, it is desirable to maximize the occurrence of crafting an item $B$, which depends on possessing the proper quantities of material. To construct 1 $B$ one needs 1 of ingredient $L$ and 2 of ingredient $S$. In addition, one can convert 1 $L$ to 4 $S$. Given this information, my goal is to determine a number $T$ of $L$ to convert to $S$. For example, if I had 3 $L$, the optimal $T$ would be 1, since then the ratio between $S$ and $L$ would be exactly 2, as desired. If I started with $l$ initial $L$ and $s$ initial $S$ (with all integer variables), then the relationship should be as follows:
$$2L = S implies\ 2(l-T)=s+4T \ 2l-2T=s+4T \ 2l-s=6T \ T=frac{2l-s}{6}
$$
I ran into some trouble refining this, because it returns non-integers for some values which are useless in-game. I've tested
$$begin{equation}T=left|frac{2l-s}{6}right|end{equation}$$
with an iterative process and determined through trial that this formula returns the correct result for $l,s in {1,2,3,4}$. This doesn't seem to be a rigorous or complete solution. In addition, this formula fails when $l=1$ and $s=6$, and possibly elsewhere.


I've supposed that this fails when a negative result is generated, because one cannot convert backwards (although the sentiment is nice.) Given the rounding (unless I improperly evaluate $|x|$), this occurs when
$$frac{2l-s}{6} leq frac{-1}{2} \ 2l-s leq -3 \ 2l leq s-3 \ 2l+3 leq s$$
This formula could then be represented as
$$T(l,s)= begin{cases}
0 & 2l+3leq s \
left|frac{2l-s}{6}right| & 2l+3>s
end{cases}
$$
This feels inelegant and non-rigorous. Is there a more general formula for this problem? Also, how could I make the conclusions about the rounding more rigorous? I have the same question for my definition of the cases in which the main formula fails, i.e. when $T=0neq |(2l-s)/6|$.

We start the game having $l$ units of material $L$ and $s$ units of $S$. After converting $T$ items of $L$ we have $l-T$ items of $L$ left and $s+4T$ items of $S$. With all this material we can obtain $minleft(l-T,frac{s+4T}2right)$ of $B$. Since crafting $B$ is the only goal we want every item of $L$ to be used. Thus we have the equality:
$$l-T=minleft(l-T,frac{s+4T}2right)$$ It is known than $min(x,y)=frac{x+y-|x-y|}2$. So we have $$l-T=frac{l+frac{s}2+T-left|l-frac{s}2-3Tright|}2$$ which simplifies to $$left|l-frac{s}2-3Tright|=3T+frac{s}2-l$$ This means that $$l-frac{s}2-3Tleq0$$ or $$Tgeqfrac{2l-s}6$$ And since $Tinmathbb{N}_0$ we get $$T=maxleft(0;leftlfloorfrac{2l-s}6rightrfloor+1right)$$ where $lfloorcdotrfloor$ is floor function.