1. **State the problem:** We want to find the point $x$ on the line between two smokestacks, $d$ miles apart, where the concentration $S$ of soot deposits is minimized. The concentration is given by $$S = \frac{c}{x^2} + \frac{k}{(d - x)^2}$$ with $k = 9c$. 2. **Substitute $k = 9c$ into the formula:** $$S = \frac{c}{x^2} + \frac{9c}{(d - x)^2} = c \left( \frac{1}{x^2} + \frac{9}{(d - x)^2} \right)$$ Since $c > 0$, minimizing $S$ is equivalent to minimizing $$f(x) = \frac{1}{x^2} + \frac{9}{(d - x)^2}$$ 3. **Find the derivative $f'(x)$ to locate critical points:** $$f'(x) = -2 \frac{1}{x^3} + 18 \frac{1}{(d - x)^3}$$ 4. **Set $f'(x) = 0$ to find critical points:** $$-\frac{2}{x^3} + \frac{18}{(d - x)^3} = 0$$ Rearranged: $$\frac{18}{(d - x)^3} = \frac{2}{x^3}$$ Cross-multiplied: $$18 x^3 = 2 (d - x)^3$$ Simplify: $$9 x^3 = (d - x)^3$$ 5. **Take cube roots:** $$\sqrt[3]{9} x = d - x$$ 6. **Solve for $x$:** $$x + \sqrt[3]{9} x = d$$ $$x (1 + \sqrt[3]{9}) = d$$ $$x = \frac{d}{1 + \sqrt[3]{9}}$$ 7. **Interpretation:** The point $x_{min}$ that minimizes the concentration is $$x_{min} = \frac{d}{1 + \sqrt[3]{9}}$$ This is the distance from the first smokestack where the soot concentration is lowest. **Final answer:** $$x_{min} = \frac{d}{1 + \sqrt[3]{9}}$$ miles