Isn't it impossible?

$Let~f(n)=\dfrac {n^2} {500+3n^3}.\\ \therefore~f'(n)=\dfrac{2n(500+3n^3)-n^2*(9n^2)}{.......}\\ For~f'_{ext},~~ f'=0.\\ \implies~ n=0~~~~~or~~~~~2(500+3n^3)-n*(9n^2)=0.\\ \therefore~For~ max~2(500+3n^3)=n*(9n^2)\\ So~n=\sqrt[3]{\dfrac {1000}3}.\\ f \left (\sqrt[3]{\dfrac {1000}3}\right )=\Huge~~\color{#D61F06}{0.03205}.$

Simply find the extrema by differentiating......and then find the integral value of n closest to the critical point......!!