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Be the function $$\begin{array}{lrll}f:&(0,+\infty)&\longrightarrow&\mathbb{R}\\&x&\longmapsto&f(x)=3x^{3}+\dfrac{A}{x^{3}}\end{array}$$ where $ A $ is a positive constant. Determine the lowest value of $ A $ such that $ f (x) \geq20 $ for all $ x> 0 $

It seems to me that I have to limit the function to values ​​greater than or equal to 20, so, I suppose I need to optimize the function, right? that is, take first derivative and choose the appropriate minimum.

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  • $\begingroup$ Yes, with appropriate considerations for what happens when $x \to 0$ and $x \to \infty$ this is a very good idea. Can you do this or do you need help? $\endgroup$
    – Ingix
    Commented Aug 27, 2018 at 22:29
  • $\begingroup$ I would like to give some clues maybe ... because, the analysis in the extremes of the interval I did not understand you very well. $\endgroup$
    – Santiago Seeker
    Commented Aug 27, 2018 at 22:35
  • $\begingroup$ hey did you understand my solution ?? or you are looking to solve it by another method $\endgroup$
    – Deepesh Meena
    Commented Aug 27, 2018 at 23:03

2 Answers 2

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You can use the A.M. G.M. method because here $x^3$ and $\frac{A}{x^3}$ are positive.

$$\frac{3x^{3}+\dfrac{A}{x^{3}}}{2}\ge (3A)^{\frac{1}{2}} $$ $$3x^{3}+\dfrac{A}{x^{3}}\ge 2(3A)^{\frac{1}{2}} $$ thus $$2(3A)^{1/2}\ge20$$ $$3A\ge 100$$ $$A\ge \frac{100}{3}$$

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  • $\begingroup$ What's the meaning or what is A.M.G.M. method? $\endgroup$
    – Santiago Seeker
    Commented Aug 27, 2018 at 22:53
  • $\begingroup$ Arithmetic mean and the geometric mean, of course, you can solve it by differentiating the function and finding the point where $f'(x)=0$ and so on $\endgroup$
    – Deepesh Meena
    Commented Aug 27, 2018 at 22:53
  • $\begingroup$ I just derived and got that $ A = 3x^{6} $ but I got stuck there. $\endgroup$
    – Santiago Seeker
    Commented Aug 27, 2018 at 23:09
  • $\begingroup$ hey you are going in the right way $x^3=\sqrt{A/3}$ put in the equation $f(x)=3\sqrt{A/3}+\frac{A}{\sqrt{A/3}}=2\sqrt{A3}$ also there was a typo in my solution check out the updated solution of mine $\endgroup$
    – Deepesh Meena
    Commented Aug 27, 2018 at 23:12
  • $\begingroup$ Thank you very much!! excuse a question ... as weapons directly the inequality, $ \frac{f (x)}{2} \geq\sqrt{3A} $ the next step if I understand $\endgroup$
    – Santiago Seeker
    Commented Aug 27, 2018 at 23:19
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Using algebra.

Consider the function $$f(x)=3x^{3}+\dfrac{A}{x^{3}}$$ $$f'(x)=9 x^2-\frac{3 A}{x^4}$$ $$f''(x)=\frac{12 A}{x^5}+18 x$$ The first derivative cancels when $$9 x^2-\frac{3 A}{x^4}=0 \implies x_*=\left(\frac A 3 \right) ^\frac 16$$ For this value $$f(x_*)=2 \sqrt{3A}$$ and the second derivative test shows that $x_*$ corresponds to a minimum. So, you want $2 \sqrt{3A} \geq 20$; then $A$.

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