Let $f(a) = \frac{13+a}{3a+7}$ where $a$ is restricted to positive integers. What is the maximum value of $f(a)$?

Question

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!

Answer 1

$f'(a)=-\frac{32}{(3a+7)^2}<0$ so $f$ is strictly decreasing for $a>-\frac73$. So $f(1)>f(2)>f(3)>...$

Answer 2

\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 $f(1) = \dfrac 75$

Answer 3

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.

Answer 4

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:

1. Translated left by $7/3$
2. Then stretched vertically by $32$
3. 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 $0$.