Answer
$\frac{1}{2}$
Work Step by Step
Step 1. Assume this positive number is $x\gt0$; define the function $y=\frac{1}{x}+4x^2$ and we need to find its minimum.
Step 2. Take the derivative of the function to get $y'=8x-\frac{1}{x^2}$; letting $y'=0$, we have $x^3=\frac{1}{8}$ and $x=\frac{1}{2}$, which gives $y=2+4(\frac{1}{2})^2=3$
Step 3. Check $y''=8+\frac{2}{x^3}\gt0$; thus we know the region is concave up and the value above is a minimum.
