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Find the positive number x such that the sum of x and its reciprocal is as small as possible.

I'm having a bit of an issue with this one. The answer in my textbook says x=1, but I can't figure out how to get it.

I assumed that the formula to be optimized is $Q=x+\frac{1}{x}$. Taking the derivative of this gets me $Q'=-\frac{1}{x^2}$ and $Q''=\frac{2}{x^3}$

The only point where a critical value can exist on $Q'$ and $Q''$ is $x=0$... but that's not right.

Where did I go wrong here?

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Recalculate the derivative for $Q$ and you may be able to figure out what went wrong. –  Clayton Dec 11 '12 at 0:50

3 Answers 3

up vote 5 down vote accepted

We restrict ourselves to $x>0$ as in the question. Differentiating $Q(x)=x+\frac{1}{x}$ gives us $Q'(x)=1-\frac{1}{x^2}$. Setting $Q'(x)=0$, we get $x=1$. Differentiating again we have $Q''(x)=\frac{2}{x^3}$ so that $Q''(1)>0$. This gives us a minimum at $x=1$.

For another method, observe that $(x+\frac{1}{x})^2-4=(x-\frac{1}{x})^2\geq 0$ so that $Q(x)\geq 2=1+\frac{1}{1}$.

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OK, so I just miscalculated my first derivative.

$$Q'=\frac{x^2-1}{x^2}$$

so $Q'=0$ when $x=\pm1$. But since the question restricts it to positive numbers, then $x=1$ it is...

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Note also that $Q'(x)=1-\frac1{x^2}<0$ when $0<x<1$ and $Q'(x)>0$ when $x>1$, so the critical point at $x=1$ really is a minimum. (Note also that if you wait a couple of days, you can accept your own solution.) –  Brian M. Scott Dec 11 '12 at 1:07

This is not exactly an answer to your question, since it does not use calculus. Note that $$x+\frac{1}{x}=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2.$$ The expression on the right is minimized when $\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2$ is as small as possible. But a square can never be negative, so the minimum is reached when $\sqrt{x}-\dfrac{1}{\sqrt{x}}=0$, that is, when $\sqrt{x}=1$.

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