This problem has left me going in circles, since I don't know how to relate the number of coins and denominations to obtain a maximum quantity. Does this problem requires the use of derivatives?. If not, then how can I solve it?
The problem is as follows:
At a vending machine factory a group of technicians have been tasked to build a new beverages dispensing machine with an upgraded cash accepting system. This mechanism can accept coins of $20$ cents, $50$ cents and $1$ dollar, $2$ dollars and $5$ dollars. After the initial run, the vending machine collected a total of $100$ dollars. Assuming that in this currency all quantities stated are coins and $100$ cents is $1$ dollar. What is the maximum quantity of coins the machine can have in its money chest if at least the machine received five coins of each denomination?.
The alternatives given in my book are:
$\begin{array}{ll} 1.&294\,\textrm{coins}\\ 2.&298\,\textrm{coins}\\ 3.&308\,\textrm{coins}\\ 4.&316\,\textrm{coins}\\ 5.&306\,\textrm{coins}\\ \end{array}$
Normally I would try to show my attempt into solving this problem, but in this particular situation I don't know how to proceed from the very beginning.
The only thing I could come up with was to build a system of $1\times 1$ equation where it would be like this:
$0.2\left(x+5\right)+0.5\left(x+5\right)+1\left(x+5\right)+5\left(x+5\right)=100$
Therefore:
$\left(x+5\right)\left(0.2+0.5+1+5\right)=100$
$\left(x+5\right)=\frac{100}{6.7}=\frac{1000}{67}$
$x=\frac{1000}{67}-5=\frac{5000-335}{67}=\frac{4665}{67}$
Needless to say that my answer doesn't get any close of what the alternatives given are. Therefore I'm requesting assistance with this particular problem. What would be the method to choose or how can I solve this riddle?.
I'm often confused whether if derivatives should apply in this situation?. Can somebody help me with an answer for this?. Something which could help me a lot it is to know if there exist a graphical approach?.