+118723 Hi Dimkorak,
Sorry I do not speak spanish but maybe you can make sense of this.
Maybe if you ask questions in Spanish someone else might be able to interpret for us.
Lo siento no hablo a español pero tal vez puede hacer sentido de esto.
Tal vez si preguntas en Español alguien podría ser capaz de interpretar para nosotros.
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Find the number in the interval [1/3, 2], such that the sum of the number and its reciprocal are a maximum.
Let the number be x
\(S=x+\frac{1}{x}\\ S=x+x^{-1}\\ \frac{dS}{dx}=1-1x^{-2}\\ \frac{dS}{dx}=1-\frac{1}{x^2}\\\)
Stat point when S'=0
\(0=1-\frac{1}{x^2}\\ \frac{1}{x^2}=1\\ x=\pm1\qquad but\quad 1/3\le x\le2 \qquad so\\ x=1 \)
\(\frac{d^2S}{dx^2}=2x^{-3}\\ \text{when x=1}\\ \frac{d^2S}{dx^2}=2>0\qquad so\\ \text{There is a minimum at x=1}\)
so the maximum must occur at one of the end points.
\(When \;\;x=1/3 \qquad S=3\frac{1}{3}\\ When\;\;x=2 \qquad \;\;\; S=2\frac{1}{2}\\~\\ \)
So the maximum will sum of the number and its reciprocal will be when the number is \(\frac{1}{3}\)
Here is a graph that you should try and make sense of.
https://www.desmos.com/calculator/zxw8w8frr9
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