The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in$^3$. Find the minimum possible sum of the three dimensions.
$\text{(A) }36 \qquad \text{(B) }38  \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$
Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \cdot 7 \cdot 11 \cdot 13$. The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$, so our answer is $11+13+14=\boxed{\textbf{(B) } 38}$