To solve the problem, we need to maximize the area of the rectangular lot, given the constraints on the fencing material.

Step 1: Define variables

• Let l be the length of the rectangular lot.
• Let w be the width of the rectangular lot.

We are told that:

• Juan will fence three sides of the lot: two widths (both sides perpendicular to the wall) and one length (the side opposite the house wall).
• Additionally, there will be a partition along the length, so we have another length to fence.

This means the total fencing used is: $2w + 2l = 117 \text{ feet}.$

Step 2: Express $w$ in terms of $l$

Rearranging the equation $2w + 2l = 117$ to isolate $w$, we get: $w = \frac{117 - 2l}{2}.$

Step 3: Write the area in terms of $l$

The area $A$ of the rectangular lot is given by: $A = l \cdot w.$

Substitute the expression for $w$ from the previous step: $A = l \cdot \frac{117 - 2l}{2}.$ Simplify the expression: $A = \frac{l(117 - 2l)}{2}.$ $A = \frac{117l - 2l^2}{2}.$ $A = 58.5l - l^2.$

Step 4: Maximize the area

To maximize the area, we take the derivative of $A$ with respect to $l$ and set it equal to zero: $\frac{dA}{dl} = 58.5 - 2l.$

Set the derivative equal to zero and solve for $l$: $58.5 - 2l = 0,$ $2l = 58.5,$ $l = 29.25 \text{ feet}.$

Step 5: Verify that this is a maximum

The second derivative of $A$ is: $\frac{d^2A}{dl^2} = -2.$ Since the second derivative is negative, this confirms that $l = 29.25$ feet gives a maximum area.

Final Answer:

The length $l$ that maximizes the area is 29.2 feet (rounded to the nearest tenth).

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

Here are 5 follow-up questions to expand on the topic:

1. What would the width of the lot be for the maximum area?
2. How would the maximum area change if Juan had 150 feet of fencing material?
3. What is the maximum possible area of the lot?
4. How would the problem change if the partition was along the width instead of the length?
5. How could we approach this problem if the fencing cost per foot varied for different sides?

Tip: Whenever you're optimizing an area or volume with constraints, try expressing all variables in terms of one variable and use calculus to find maximum or minimum values.