I am trying to figure out a mixing problem and I'm stuck because it seems to have two levels. I'm going to write a simpler form here of the problem I am working on.
Say that we have 60 pounds of a substance that contains 25% A, 30% B and 40% C. Say again that I have 35 pounds of a substance that contains 15% A, 10% B, and 10% C. Finally say that I have 100 pounds of substance with 5% A, 20% B, 35% C.
How can I mix these three substances so as to obtain a mixture with 20% A, 20% B, and 20% C.
Obviously a problem like this may not even be feasible. What field of mathematics is this anyway? I'm trying to figure this out for work so I need to get a grasp conceptually of what I'm doing here.
Let vectors in $\mathbb{R}^4$ describe the distribution of the constituents $A$, $B$, $C$ and other $O$. For example, the distribution of substance 1 is
$$s_1=\left(\begin{array}{c}0.25\\0.3\\ 0.4\\0.05\end{array}\right).$$
Now we want $n_i$ units of each substance $i$ to produce a distribution:
$$s=\left(\begin{array}{c}0.2\\0.2\\ 0.2\\0.4\end{array}\right).$$
Now if we have $n_i$ of each substance in total we have a mix composed of $N=n_1+n_2+n_3$ units and the amount of stuff in each is
$$(n_1+n_2+n_3)\left(\begin{array}{c}0.2\\0.2\\ 0.2\\0.4\end{array}\right)=\left(\begin{array}{c}0.2(n_1+n_2+n_3)\\0.2(n_1+n_2+n_3)\\ 0.2(n_1+n_2+n_3)\\0.4(n_1+n_2+n_3)\end{array}\right).$$
Now if we break this total down into the three substances what we see we want $$ \begin{align} n_1\left(\begin{array}{c}0.25\\0.3\\ 0.4\\0.05\end{array}\right)+n_2\left(\begin{array}{c}0.15\\0.1\\ 0.1\\0.65\end{array}\right)+n_3\left(\begin{array}{c}0.05\\0.2\\ 0.35\\0.4\end{array}\right) \\=\left(\begin{array}{c}0.25n_1\\0.3n_1\\ 0.4n_1\\0.05n_1\end{array}\right)+\left(\begin{array}{c}0.15n_2\\0.1n_2\\ 0.1n_2\\0.65n_2\end{array}\right)+\left(\begin{array}{c}0.05n_3\\0.2n_3\\ 0.35n_3\\0.4n_3\end{array}\right) \\=\left(\begin{array}{c}0.25n_1+0.15n_2+0.05n_3\\0.3n_1+0.1n_2+0.2n_3\\ 0.4n_1+0.1n_2+0.35n_3\\0.05n_1+0.65n_2+0.4n_3\end{array}\right)&\overset{!}{=}\left(\begin{array}{c}0.2(n_1+n_2+n_3)\\0.2(n_1+n_2+n_3)\\ 0.2(n_1+n_2+n_3)\\0.4(n_1+n_2+n_3)\end{array}\right). \end{align}.$$
Now this is four equations in three unknowns:
$$\begin{align}0.05n_1-0.05n_2-0.15n_3&=0 \\0.1n_1-0.1n_2+0n_3&=0 \\ 0.2n_1-0.1n_2+0.15n_3&=0 \\ -0.35n_1+0.45n_2+0.2n_3&=0. \end{align}$$
Can you solve this linear system? There is every chance that there are no non-trivial solutions as you have more equations than unknowns.
The trivial solution is $n_1=n_2=n_3=0$ which isn't very interesting!