How to do the computation to find the number of coins in a chest if there are different denominations? 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?.

You seem to have forgotten about the $2$ dollar coin. Since we know that there are at least five coins of each value, we can subtract:
$$5 \cdot (0.2 + 0.5 + 1 + 2 + 5) = 5 \cdot 8.7 = 43.5$$
We thus need to achieve the highest number of coins that achieve a sum of $100 - 43.5 = 56.5$. Most of the times, we can achieve this by assigning the lowest value first. Note that, had the remaining amount been a multiple of $0.2$, we could have simply divided this amount by $0.2$ to achieve the highest number of coins. However, since this is not the case, we first need to subtract one coin of $0.5$ to find us left with a total sum of $56$. As such, the maximum number of coins equals:
$$5 \cdot 5 + 1 + \frac{56}{0.2} = 306$$
