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 $p \leq x\leq15.$
Since $0 < p \le x \le 15,$ the absolute values simplify to \[f(x) = (x-p) - (x-15) - (x-p-15) = -x+30.\]The value of this expression is minimized when $x=15,$ giving $-15+30=\boxed{15}.$