Is it possible to maximize $\frac{3t^2}{t^3+4}$ (where $t>0$) without taking derivative? To find the maximum of $f(t)=\dfrac{3t^2}{t^3+4}$ (for $t>0$) we can simply equate the derivative with zero,
$$f'(t)=0\Rightarrow 6t(t^3+4)-3t^2(3t^2)=0\Rightarrow -3t^4+24t=0\Rightarrow t=2$$
And $f_{max}=f(2)=1$.
I'm wondering is it possible to find the maximum without taking derivative? I'm eager to see other methods to maximize the function.

You can use AM-GM as follows:
\begin{eqnarray*} \frac{3t^2}{t^3+4}
& = & \frac{3t^2}{\frac 12 t^3 + \frac 12 t^3 +4} \\
& \stackrel{AM-GM}{\leq} &  \frac{3t^2}{3\sqrt[3]{\frac 12 t^3 \cdot \frac 12 t^3 \cdot 4}} \\
& = & 1
\end{eqnarray*}
Equality holds iff
$$\frac 12 t^3 = 4 \Leftrightarrow t= 2$$
