Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$? Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$?
I tried graphing but it didn't help.  Could anyone answer?  Thanks!
 A: \begin{align}
   f(a)
   &= \dfrac{a+13}{3a+7} \\
   &= \dfrac 13 \dfrac{3a+39}{3a+7} \\
   &= \dfrac 13 \left(1 + \dfrac{32}{3a+7} \right) \\
\end{align}
This implies that $f(a)$ is strictly decreasing for positive integers.
Hence the maximum value must be $f(1) = \dfrac 75$
A: $f'(a)=-\frac{32}{(3a+7)^2}<0$ so $f$ is strictly decreasing for $a>-\frac73$. So $f(1)>f(2)>f(3)>...$
A: I plotted the graph and it show two branches, as you are limiting it to positive values of $a$ it is the right hand branch which is relevant. This shows as crossing the $y-axis$ at the point where $a=0$ and thereafter the curve is decreasing towards the $x-axis$. so putting $a=0$ gives the max value as $13/7$. If you further limit it to positive integer values of $x$ then this occurs for $x=1$ to give a max value of 14/10. 
A: Expanding on @StevenGregory's answer, which is more than satisfactory:
Since $f(a)=\frac{1}{3}+\frac{32}{a+7/3}$, we can see that it's got the same graph as that of $y=\frac{1}{a}$ but:


*

*Translated left by $7/3$

*Then stretched vertically by $32$

*Then translated vertically by $1/3$


These transformations don't change the fact that the right branch is decreasing except that the right branch now starts at $-\frac{7}{3}$, instead of $0$.
