1. **State the problem:** We are given that the surface area $S$ of a melting snowball decreases at a rate of $\frac{dS}{dt} = -4.2$ cm$^2$/min. We need to find the rate at which its diameter $D$ decreases when $D = 15$ cm. 2. **Recall the formula for the surface area of a sphere:** $$S = 4\pi r^2$$ where $r$ is the radius of the sphere. 3. **Relate diameter and radius:** $$D = 2r \implies r = \frac{D}{2}$$ 4. **Express $S$ in terms of $D$:** $$S = 4\pi \left(\frac{D}{2}\right)^2 = 4\pi \frac{D^2}{4} = \pi D^2$$ 5. **Differentiate $S$ with respect to time $t$:** $$\frac{dS}{dt} = \frac{d}{dt}(\pi D^2) = 2\pi D \frac{dD}{dt}$$ 6. **Plug in known values and solve for $\frac{dD}{dt}$:** Given $\frac{dS}{dt} = -4.2$ cm$^2$/min and $D = 15$ cm, $$-4.2 = 2\pi (15) \frac{dD}{dt}$$ 7. **Isolate $\frac{dD}{dt}$:** $$\frac{dD}{dt} = \frac{-4.2}{2\pi \times 15} = \frac{-4.2}{30\pi}$$ 8. **Simplify the fraction:** $$\frac{dD}{dt} = -\frac{4.2}{30\pi} = -\frac{\cancel{4.2}}{\cancel{30}\pi} = -\frac{0.14}{\pi} \approx -0.0446$$ 9. **Interpretation:** The diameter decreases at approximately $0.0446$ cm/min when the diameter is 15 cm. **Final answer:** $$\boxed{\frac{dD}{dt} \approx -0.0446 \text{ cm/min}}$$