To determine the time when the rate of flow is greatest, we need to find the maximum value of the derivative of the volume function $V(t)$. The rate of flow is given by the derivative of the volume with respect to time, $V'(t)$.

Given the volume function:

$V(t) = \frac{1}{300} \left( 2t^3 - \frac{t^4}{4} \right)$

Step 1: Find the derivative $V'(t)$

We calculate the derivative $V'(t)$ with respect to $t$:

$V'(t) = \frac{d}{dt} \left( \frac{1}{300} \left( 2t^3 - \frac{t^4}{4} \right) \right)$

$V'(t) = \frac{1}{300} \left( 6t^2 - t^3 \right)$

$V'(t) = \frac{1}{300} \cdot t^2 \left( 6 - t \right)$

Thus, the rate of flow is:

$V'(t) = \frac{t^2(6 - t)}{300}$

Step 2: Find the critical points of $V'(t)$

To maximize the rate of flow, we need to find the critical points by setting the second derivative $V''(t)$ equal to zero.

Let's differentiate $V'(t)$ to find $V''(t)$:

$V''(t) = \frac{d}{dt} \left( \frac{t^2(6 - t)}{300} \right)$

Expanding the product:

$V''(t) = \frac{d}{dt} \left( \frac{6t^2 - t^3}{300} \right)$

$V''(t) = \frac{1}{300} \left( 12t - 3t^2 \right)$

$V''(t) = \frac{1}{300} \cdot 3t(4 - t)$

Step 3: Solve for the critical points

Set $V''(t) = 0$:

$3t(4 - t) = 0$

This gives us:

$t = 0 \quad \text{or} \quad t = 4$

Since $t = 0$ corresponds to the start of the process, the maximum rate of flow occurs at $t = 4$ hours.

Conclusion

The rate of flow was greatest at 4 hours.

Would you like more details or have any questions about this solution?

Here are 5 related questions to explore:

1. How do you determine if a critical point is a maximum or a minimum?
2. What is the physical significance of the second derivative in this context?
3. How would the solution change if the pond emptied in 8 hours instead of 6 hours?
4. Can the maximum rate of flow occur at a boundary point?
5. How would the problem change if the volume function had an additional constant term?

Tip: The second derivative test is a quick way to determine whether a critical point is a maximum, minimum, or inflection point.