Answer:
--- Objective function
Interval = 
Step-by-step explanation:
Given
Represent the number with x
The required sum can be represented as:

Hence, the objective function is:

To get the the interval, we start by differentiating w.r.t x
<em>Using first principle, this gives:</em>

Equate S'(x) to 0 in order to solve for x

Subtract 1 from both sides


Multiply both sides by -1

Cross Multiply


Take positive square root of both sides because x is positive


Representing x using interval notation, we have
Interval = 
To get the smallest sum, we substitute 1 for x in 



<em>Hence, the smallest sum is 2</em>