1. **State the problem:** We are given a cube whose side length is increasing at a rate of 12 m/s. We need to find the rate at which the volume of the cube is increasing when the side length is 4 cm. 2. **Identify variables and given data:** Let $s$ be the side length of the cube (in meters), and $V$ be the volume of the cube (in cubic meters). Given: - Rate of change of side length: $\frac{ds}{dt} = 12$ m/s - Side length at the instant: $s = 4$ cm = 0.04 m (converted to meters) 3. **Formula for volume of a cube:** $$V = s^3$$ 4. **Differentiate volume with respect to time $t$ to find rate of change of volume:** $$\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$$ 5. **Substitute known values:** $$\frac{dV}{dt} = 3 \times (0.04)^2 \times 12$$ 6. **Calculate:** $$\frac{dV}{dt} = 3 \times 0.0016 \times 12 = 0.0576$$ 7. **Interpretation:** The volume of the cube is increasing at a rate of $0.0576$ cubic meters per second when the side length is 4 cm. **Final answer:** $$\boxed{\frac{dV}{dt} = 0.0576 \text{ m}^3/\text{s}}$$