To solve this problem, we are tasked to determine how fast the sides of the cube are increasing when the volume of the cube grows at 8 m³/min, and the side length of the cube is 6 m.

Step-by-Step Solution

1. Volume of a Cube: The volume $V$ of a cube is given by: $V = s^3$ where $s$ is the side length of the cube.

2. Differentiate with respect to time $t$: To relate the rate of change of volume $\frac{dV}{dt}$ with the rate of change of the side length $\frac{ds}{dt}$, differentiate both sides of $V = s^3$ with respect to $t$: $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}.$

3. Substitute Known Values:

   • The rate of change of volume is $\frac{dV}{dt} = 8 \, \text{m}^3/\text{min}$.
   • The side length is $s = 6 \, \text{m}$.

   Plug these values into the equation: $8 = 3(6^2) \frac{ds}{dt}.$

4. Simplify and Solve for $\frac{ds}{dt}$:

   • First, calculate $6^2 = 36$: $8 = 3(36) \frac{ds}{dt}.$
   • Simplify $3 \times 36 = 108$: $8 = 108 \frac{ds}{dt}.$
   • Solve for $\frac{ds}{dt}$ (divide both sides by 108): $\frac{ds}{dt} = \frac{8}{108}.$
   • Simplify: $\frac{ds}{dt} = \frac{2}{27}.$

5. Final Answer: The rate at which the sides of the cube are increasing is: $\frac{ds}{dt} = \frac{2}{27} \, \text{m/min}.$

Conclusion:

When the side length of the cube is 6 meters, the sides are increasing at a rate of $\frac{2}{27} \, \text{m/min}$.

Do you want me to explain this step further or clarify any part?

Related Questions:

1. How does the rate of change of the side length depend on the volume growth rate?
2. What happens to $\frac{ds}{dt}$ if $s$ increases to 10 m?
3. How would the problem change if the cube’s volume grew at $12 \, \text{m}^3/\text{min}$?
4. Can you derive a general formula for $\frac{ds}{dt}$ in terms of $\frac{dV}{dt}$ and $s$?
5. What is the geometric interpretation of this problem?

Tip:

When solving related rates problems, always differentiate the given equation with respect to time $t$, apply the chain rule carefully, and substitute known values to isolate the desired rate.