Given:

• Volume of the cube $V = s^3$, where $s$ is the side length.
• The rate of change of the volume is given by $\frac{dV}{dt} = -1584 \, \text{m}^3/\text{min}$.
• We need to find the rate of change of the surface area $S = 6s^2$ at the instant when $s = 7 \, \text{m}$.

Step-by-step solution:

1. Differentiate the volume equation with respect to $t$: $V = s^3 \quad \Rightarrow \quad \frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ Plugging in the known values: $-1584 = 3 \cdot (7^2) \cdot \frac{ds}{dt}$ $-1584 = 3 \cdot 49 \cdot \frac{ds}{dt}$ $-1584 = 147 \cdot \frac{ds}{dt}$ Solving for $\frac{ds}{dt}$: $\frac{ds}{dt} = \frac{-1584}{147} \approx -10.7755 \, \text{m/min}$

2. Differentiate the surface area equation with respect to $t$: $S = 6s^2 \quad \Rightarrow \quad \frac{dS}{dt} = 12s \frac{ds}{dt}$ Plugging in the known values: $\frac{dS}{dt} = 12 \cdot 7 \cdot (-10.7755)$ $\frac{dS}{dt} = 84 \cdot (-10.7755)$ $\frac{dS}{dt} \approx -905.142 \, \text{m}^2/\text{min}$

Final Answer:

The rate of change of the surface area of the cube at the instant when the side length is 7 meters is approximately $-905.142 \, \text{m}^2/\text{min}$.

Would you like further explanation or have any questions about this solution?

Here are some related questions for deeper understanding:

1. How is the derivative of the volume of the cube related to the side length in general?
2. Why is the rate of change of the surface area negative in this context?
3. What is the meaning of a negative rate of change for both volume and surface area in real-world terms?
4. How would the calculations change if the rate of volume change were positive instead?
5. Can you describe a scenario where the surface area change might be more important than the volume change?

Tip: Always pay attention to the signs of rates of change to understand whether a quantity is increasing or decreasing.