# 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.

• Isn't there supposed to be a $(t^3 + 4)^2$ in the quotient? 2 days ago
• @ewong I didn't write $f'(x)$. I just equated its numerator with zero. 2 days ago
• Oh. ok. Was a bit confused. 2 days ago
• Hint: when $t \gt 0\,$, by AM-GM $\;\dfrac{3}{f(t)} = t + \dfrac{4}{t^2}= \dfrac{t}{2} + \dfrac{t}{2} + \dfrac{4}{t^2} \ge \dots$
– dxiv
2 days ago
• @dxiv Thanks a lot! Your method is very interesting! 2 days ago

For $$0\leq t\leq 1$$

We have the obvious inequality :

$$f(t)=\dfrac{3t^2}{t^3+4}\leq g(t)=\dfrac{3t^2}{t^4+4}$$

But with Germain's indentity:

$$t^4+4=(t^2-2t+2)(t^2+2t+2)$$

So decomposing :

$$g(t)=\frac{3t}{4\left(t^{2}-2t+2\right)}-\frac{3t}{4\left(t^{2}+2t+2\right)}$$

But :

$$h(t)=\frac{3t}{4\left(t^{2}-2t+2\right)}$$

$$t^{2}-2t+2$$ is decreasing for $$t\in[0,1]$$ and $$3t$$ is increasing so :

$$f(t)

For $$t\geq 1$$ we have the inequality :

$$f(t)\leq \frac{4t}{4+t^{2}}$$

And as $$t^2+4-4t\geq 0$$ the inequality follows .

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 $$\frac 12 t^3 = 4 \Leftrightarrow t= 2$$