The volume of water flowing out of a pond is modelled by the equation V=1/300(2t^3-t^4/4), where t is the time in hours. If the pond takes 6 hours to empty completely, calculate the time when the rte of flow was the greatest

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.

Learn how to calculate the time when the rate of water flow from a pond was at its maximum using calculus. This problem involves finding critical points of a volume function and applying the second derivative test to determine maximum flow rate time. Suitable for advanced high school students studying calculus.

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