A calculus problem by Rahil Sehgal

$a_{n}$ will be maximized when $\dfrac{1}{a_{n}} = \dfrac{500}{n^{2}} + 3n$ is minimized. Now by the AM-GM inequality we have that

$\dfrac{500}{n^{2}} + 3n = \dfrac{500}{n^{2}} + \dfrac{3n}{2} + \dfrac{3n}{2} \ge 3\sqrt[3]{\dfrac{500}{n^{2}} \times \left(\dfrac{3n}{2}\right)^{2}} = \dfrac{9}{2}\sqrt[3]{500}$ ,

with equality holding when $\dfrac{500}{n^{2}} = \dfrac{3n}{2} \Longrightarrow n = \sqrt[3]{\dfrac{1000}{3}} \approx 6.9336$ .

Now the closest integer value is $7$ , but to confirm that this is the solution we need to check that $a_{6}$ and $a_{8}$ are less than $a_{7}$ , which indeed they are. Thus the answer in $\boxed{7}$ .

Comment: We could also use calculus. Taking $n = x$ as a continuous variable, we have that

$a'(x) = \dfrac{2x(500 + 3x^{3}) - x^{2}(9x^{2})}{(500 + 3x^{3})^{2}} = 0$ when $x = 0$ or $1000 = 3x^{3} \Longrightarrow x = \sqrt[3]{\dfrac{1000}{3}}$ ,

which is consistent with our previous result.

Not a solution but i found this in Fiitjee RSM. IIT aspirants may know this