1. **Problem Statement:** The radius $r$ of a circular oil patch is increasing at a rate of 1.2 cm/min. Find the rate at which the surface area $A$ of the patch is increasing when $r=25$ cm. 2. **Formula Used:** The surface area of a circle is given by $$A = \pi r^2$$ 3. **Differentiate with respect to time $t$:** $$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$$ 4. **Given values:** - $\frac{dr}{dt} = 1.2$ cm/min - $r = 25$ cm 5. **Substitute values into the differentiated formula:** $$\frac{dA}{dt} = 2\pi (25)(1.2)$$ 6. **Calculate:** $$\frac{dA}{dt} = 2 \times \pi \times 25 \times 1.2 = 60\pi$$ 7. **Final answer:** The surface area is increasing at a rate of $$60\pi \approx 188.5$$ cm$^2$/min when the radius is 25 cm.