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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.

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  • $\begingroup$ Isn't there supposed to be a $(t^3 + 4)^2$ in the quotient? $\endgroup$
    – ewong
    2 days ago
  • $\begingroup$ @ewong I didn't write $f'(x)$. I just equated its numerator with zero. $\endgroup$
    – Amirali
    2 days ago
  • $\begingroup$ Oh. ok. Was a bit confused. $\endgroup$
    – ewong
    2 days ago
  • 5
    $\begingroup$ Hint: when $t \gt 0\,$, by AM-GM $\;\dfrac{3}{f(t)} = t + \dfrac{4}{t^2}= \dfrac{t}{2} + \dfrac{t}{2} + \dfrac{4}{t^2} \ge \dots$ $\endgroup$
    – dxiv
    2 days ago
  • $\begingroup$ @dxiv Thanks a lot! Your method is very interesting! $\endgroup$
    – Amirali
    2 days ago

2 Answers 2

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Answer using algebraic identity :

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 :

$f(t)<h(1)=0.75$

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 .

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

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