Juan wants to fence three sides of a rectangular lot for his dog. The fourth side of the lot will be a wall of his house and therefore doesn't need fencing. Instead, he added a partition inside the rectangular lot. (see illustration below)

            

He has 115 feet of fencing material available for the three sides and the partition.

Find the width  
w
 , in feet, so that the area is a MAXIMUM.

 

Note: Round-off your answer to the nearest tenths

Let's set up the problem using the following variables:

- Let **w** be the width of the rectangular lot (the dimension perpendicular to the house wall).
- Let **L** be the length of the rectangular lot (the dimension parallel to the house wall).
- Juan needs to fence three sides of the lot, two widths and one length, plus a partition parallel to the width inside the lot.

So, the total amount of fencing material used is given by:

\[
2w + L + w = 115
\]

This simplifies to:

\[
3w + L = 115
\]

Solving for \(L\):

\[
L = 115 - 3w
\]

Now, we know that the **area** \(A\) of the rectangular lot is:

\[
A = L \times w
\]

Substitute \(L = 115 - 3w\) into the area equation:

\[
A(w) = w(115 - 3w)
\]

Simplify this:

\[
A(w) = 115w - 3w^2
\]

To find the value of \(w\) that maximizes the area, we need to take the derivative of \(A(w)\) with respect to \(w\) and set it equal to zero.

\[
\frac{dA}{dw} = 115 - 6w
\]

Set the derivative equal to zero:

\[
115 - 6w = 0
\]

Solving for \(w\):

\[
6w = 115
\]
\[
w = \frac{115}{6} \approx 19.17 \text{ feet}
\]

So, the width \(w\) that maximizes the area is approximately 19.2 feet (rounded to the nearest tenth).

### Summary:
- The width \(w\) that maximizes the area is approximately **19.2 feet**.

Would you like more details, or do you have any questions?

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Here are 5 related questions to explore:

1. What is the corresponding length \(L\) when the width is 19.2 feet?
2. How do we verify that this width gives a maximum area, not a minimum?
3. What would happen to the area if Juan had more or less fencing material?
4. How would the solution change if the partition was parallel to the length instead?
5. Can we solve for the area if the width is fixed at another value, like 15 feet?

**Tip**: When maximizing or minimizing, always check that the critical point leads to the correct type of extremum by using the second derivative!

Juan wants to fence three sides of a rectangular lot for his dog. The fourth side of the lot will be a wall of his house and therefore doesn't need fencing. Instead, he added a partition inside the rectangular lot. He has 115 feet of fencing material available for the three sides and the partition. Find the width w, in feet, so that the area is a MAXIMUM. Note: Round-off your answer to the nearest tenths.

Solve the problem of maximizing the area of a rectangular lot using 115 feet of fencing material for three sides and a partition. Learn how to set up the equations and use derivatives for optimization. Suitable for high school students in Grades 10-12.

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