jmc

We can write the problem as the inequality $w(w+15)\ge 100$. Distributing on the left-hand side, subtracting 100 from both sides, and factoring, we get $$\begin{array}{rl}w(w+15)& \ge 100\Rightarrow \\ w^2+15w-100& \ge 0\Rightarrow \\ (w-5)(w+20)& \ge 0.\end{array}$$ The roots are $w=5$ and $w=-20$. We can't have a width of -20 ft, so the smallest width possible while still having an area of at least 100 sq. ft is $\boxed{5}$ ft.