Problem:
Let f(x)=∣x−p∣+∣x−15∣+∣x−p−15∣, where 0<p<15. Determine the minimum value taken by f(x) for x in the interval p≤x≤15.
Solution:
Since 0<p≤x≤15, then ∣x−p∣=x−p,∣x−15∣=15−x, and ∣x−(p+15)∣=p+15−x. Thus
f(x)=(x−p)+(15−x)+(p+15−x)=30−x
It follows that f(x) is least when x is greatest, and that the answer is 15.
The problems on this page are the property of the MAA's American Mathematics Competitions