Chapter 4: Applications of Derivatives - Section 4.5 - Applied Optimization - Exercises 4.5 - Page 224: 29

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Step 1. Assume the positive number is $x\gt0$; then $y=x+\frac{1}{x}$ is the function we need to find the minimum. Step 2. Take the derivative of the function to get $y'=1- \frac{1}{x^2}$. Let $y'=0$; we get $x=1$ and $y=2$ for this case. Step 3. Check $y''=\frac{2}{x^3}\gt0$; we know that the region is concave up and the value above is a minimum.