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

I gave up on this problem. The solution said you can write the function as:

\[f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7} = \frac13 + \frac{32}{9a+21}.\]

I get where to go from the last step, but how did (13 + a)/(3a + 7) turn into ((1/3)(3a+7))/(3a+7)?

Coolmonkey Jun 6, 2024

#1**+1 **

Hello!

So, first, let's review. We know that

\(f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7}\) . I believe you understand this part!

Now, for the 1/3 in the second part where it starts getting confusing, let's first note that \(\frac{(ax+b)}{(x+b)} = \frac{a}{1} = a\). Since the polynomial is equivalent, we can cancel them out. That's what we are doing here. Canceling out the 3a + 7, we get the remaining 1/3.

Now, for the \(\frac{32}{9a+21}\). Let's note that \(\frac{a}{b} = \frac{a}{b} \cdot \frac{c}{c}\). We want to get rid of the 3 in the numerator, so we multiplythe numerator and denominator by 3.

We get \(\frac{32/3}{3a+7} = \frac{32/3}{3a+7} \cdot \frac{3}{3} = \frac{32}{3(3a+7)} = \frac{32}{9a+21}\)

So there, that's how we got from the second step to the third step.

Good luck from here!

I hope I answered your question!

Thanks! :)

NotThatSmart Jun 6, 2024