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!