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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)?

 Jun 6, 2024
 #1
avatar+759 
+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! :)

 Jun 6, 2024
edited by NotThatSmart  Jun 6, 2024
edited by NotThatSmart  Jun 6, 2024
 #2
avatar+129399 
+1

f(a)  =  (13 + a) / (3a + 7)

 

There is  no max value

 

At  a = 0   f(a)  =  13/7

 

As a →  inf,    f(a) →  1/3

 

See the graph here : https://www.desmos.com/calculator/reaplfkcny

 

 

cool cool cool 

 Jun 6, 2024

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