To double the range, we must find the current range, which is $28-3=25$, to then double to: $2(25)=50$. Since we do not want to change the median, we need to get a value less than $8$ (as $8$ would change the mode) for the smaller, making $53$ fixed for the larger. Remember, anything less than $3$ is not beneficial to the optimization. So, taking our optimal values of $7$ and $53$, we have an answer of $7+53=$ (D) $60$. 
Two integers are inserted into the list $3,3,8,11,28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers? (2023 AMC 8 Problems, Question \#20) 
Options:
  * $$56$$ 
  * $$57$$ 
  * $$58$$ 
  * $$60$$ 
  * $$61$$ 