• # question_answer The positive, integer k for which $\frac{{{k}^{2}}}{3{{k}^{3}}+500}$ is a maximum is A) 5                                  B) 6       C) 7          D) 8

[c]  Let $f(x)=\frac{{{x}^{2}}}{3{{x}^{3}}+500}$ and $g(x)=3x+\frac{500}{{{x}^{2}}}$ If $g(x)$is minimum, then $f(x)$is maximum $g'(x)=3-\frac{1000}{{{x}^{3}}}$$\Rightarrow$$g''(x)\frac{3000}{{{x}^{4}}}>0$ Minimum value of $g(x)$at $x={{\left( \frac{1000}{3} \right)}^{1/3}}$$={{(333.33)}^{1/3}}$
 Where $6f(6)$ $\therefore$ A $k=7$is maximum.