(c)      Use differentiation to determine the width W that gives the maximum area A for the enclosure, and show that this is a maximum. What is the
 
maximum enclosure area? Give your answers to two decimal places.


To determine the width \( W \) that gives the maximum area \( A \) for the enclosure, we will follow these steps:

1. **Differentiate the area function \( A(W) \) with respect to \( W \).**
2. **Find the critical points by setting the derivative equal to zero.**
3. **Determine whether the critical point corresponds to a maximum using the second derivative test.**
4. **Calculate the maximum area.**

### 1. Differentiate the Area Function \( A(W) \)

The area function is given by:

\[
A(W) = 10W - \frac{\pi W^2}{4}
\]

Differentiate \( A(W) \) with respect to \( W \):

\[
\frac{dA}{dW} = \frac{d}{dW} \left(10W - \frac{\pi W^2}{4}\right)
\]

The derivative is:

\[
\frac{dA}{dW} = 10 - \frac{\pi W}{2}
\]

### 2. Find the Critical Points

To find the critical points, set the derivative equal to zero:

\[
10 - \frac{\pi W}{2} = 0
\]

Solve for \( W \):

\[
\frac{\pi W}{2} = 10
\]

\[
W = \frac{20}{\pi}
\]

So, the critical point is:

\[
W = \frac{20}{\pi} \approx 6.37 \text{ meters}
\]

### 3. Second Derivative Test

To determine whether this critical point corresponds to a maximum, we use the second derivative test.

Differentiate \( \frac{dA}{dW} \) again with respect to \( W \):

\[
\frac{d^2A}{dW^2} = \frac{d}{dW}\left(10 - \frac{\pi W}{2}\right)
\]

The second derivative is:

\[
\frac{d^2A}{dW^2} = -\frac{\pi}{2}
\]

Since \( \frac{d^2A}{dW^2} = -\frac{\pi}{2} \) is negative, the function \( A(W) \) is concave down at \( W = \frac{20}{\pi} \), meaning that this critical point is indeed a maximum.

### 4. Calculate the Maximum Area

Finally, substitute \( W = \frac{20}{\pi} \) into the original area formula to find the maximum area:

\[
A_{\text{max}} = 10W - \frac{\pi W^2}{4}
\]

Substitute \( W = \frac{20}{\pi} \):

\[
A_{\text{max}} = 10\left(\frac{20}{\pi}\right) - \frac{\pi \left(\frac{20}{\pi}\right)^2}{4}
\]

Simplify the expression:

\[
A_{\text{max}} = \frac{200}{\pi} - \frac{400}{4\pi} = \frac{200}{\pi} - \frac{100}{\pi} = \frac{100}{\pi}
\]

Thus:

\[
A_{\text{max}} \approx \frac{100}{3.1416} \approx 31.83 \text{ square meters}
\]

### Final Answers:

- The width \( W \) that gives the maximum area is approximately \( 6.37 \) meters.
- The maximum area is approximately \( 31.83 \) square meters.

### Do you want details or have any questions?

### Related Questions:
1. How would the maximum area change if the coefficient of \( W \) in the linear term was different?
2. Can we use this method to find the minimum area instead?
3. What is the significance of the second derivative in determining maxima or minima?
4. How does the area function behave for values of \( W \) larger or smaller than \( \frac{20}{\pi} \)?
5. What if the shape enclosed is different, how would the formula for maximum area change?
6. Can we graphically represent the area function and its maximum point?
7. How does the maximum area relate to the total perimeter of the enclosure?
8. How can this approach be generalized to optimize other functions, like volume or cost?

### Tip:
When maximizing or minimizing a function using calculus, always check the nature of the critical points using the second derivative to ensure whether it's a maximum or minimum.

Learn how to maximize the area of a rectangle using differentiation in calculus. Step-by-step solution to find the width that maximizes the area, with the maximum area calculated to two decimal places. Suitable for advanced high school students.

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