1. **State the problem:** We have a conical tank with radius $r$ and height $h=12$ feet. Brine flows in at a rate of $\frac{dV}{dt} = 3$ cubic feet per minute. We want to find how fast the radius $r$ is increasing, i.e., $\frac{dr}{dt}$, when $r=2$ feet. 2. **Relevant formulas and relationships:** The volume $V$ of a cone is given by: $$V = \frac{1}{3} \pi r^2 h$$ Since the cone's shape is fixed, the radius and height are related by similar triangles: $$\frac{r}{h} = \frac{5}{12} \implies h = \frac{12}{5} r$$ 3. **Express volume $V$ in terms of $r$ only:** Substitute $h = \frac{12}{5} r$ into the volume formula: $$V = \frac{1}{3} \pi r^2 \left(\frac{12}{5} r\right) = \frac{1}{3} \pi \frac{12}{5} r^3 = \frac{12 \pi}{15} r^3 = \frac{4 \pi}{5} r^3$$ 4. **Differentiate volume with respect to time $t$:** $$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4 \pi}{5} r^3 \right) = \frac{4 \pi}{5} \cdot 3 r^2 \frac{dr}{dt} = \frac{12 \pi}{5} r^2 \frac{dr}{dt}$$ 5. **Solve for $\frac{dr}{dt}$:** Given $\frac{dV}{dt} = 3$ and $r=2$, substitute these values: $$3 = \frac{12 \pi}{5} (2)^2 \frac{dr}{dt} = \frac{12 \pi}{5} \cdot 4 \frac{dr}{dt} = \frac{48 \pi}{5} \frac{dr}{dt}$$ Rearranging: $$\frac{dr}{dt} = \frac{3 \cdot 5}{48 \pi} = \frac{15}{48 \pi} = \frac{5}{16 \pi}$$ 6. **Final answer:** The radius is increasing at a rate of $$\boxed{\frac{5}{16 \pi} \text{ feet per minute}}$$ when the radius is 2 feet.