$a,b,c$ are digits. İf $a+2b+3c=40$ then what is the largest value for $a+b+c$? I am finding it hard to understand the method and the logic behind this. 
I read that we have to give the one with the smallest coefficient the biggest value ,why is this . ?
 A: 
I read that we have to give the one with the smallest coefficient the biggest value; why is this?

Let's say you are going to market to buy fish.  Some fish cost \$1, some cost \$2, others cost \$3.  You have \$40 and want to buy the most fish.  How many of each should you buy?  Clearly you should buy 40 of the one-dollar fish.  You could swap two of the one-dollar fish for one two-dollar fish; that would result in the same amount paid, but would result in one less fish.
A: Assume you have found an optimal triple $(a,b,c)$ with $a,b,c\in\{0,\ldots,9\}$. 
Note that $a+2b+3c=40$ implies $3c=40-a-2b\ge 40-9-18=13$ and so $c>4$.
Also, $2b=40-a-3c\ge 40-9-27=4$ and so $b\ge 2$.
If $a<8$ in our optimal solution, then (as $b\ge 2$ and so $b-1$ is still a digit), then $(a+2,b-1,c)$ is a valid triple and has larger target sum, contradicting optimality.
We conclude that $a\ge 8$ in any optimal triple.
Similarly, if $b<7$, then $(a,b+3,c-2)$ is an improvement. We conclude that $b\ge 7$.
In particular, $a$  and $b$ are not far apart: $|a-b|\le 2$. then from  $a-b=40-3b-3c=1+3\cdot(13-b-c)$, we conclude that $a-b=1$ and also that $13-b-c=0$. This makes $a+b+c=13+a$, so we certainly want to make $a$ as large as possible, namely $a=9$, leading to  the maximal target sum $a+b+c=13+a=22$, where by the way $b=a-1=8$ and $c=13-b=5$.
