1. **Problem statement:** Santa wants to build a rectangular enclosure using fencing and wants to find the dimensions that maximize the area of the enclosure. 2. **Formula and rules:** Let the length be $L$ and the width be $W$. The perimeter $P$ is fixed because the amount of fencing is fixed. The perimeter formula is: $$P = 2L + 2W$$ The area $A$ of the rectangle is: $$A = L \times W$$ 3. **Express one variable in terms of the other:** From the perimeter formula, $$2L + 2W = P \implies L + W = \frac{P}{2} \implies W = \frac{P}{2} - L$$ 4. **Substitute into the area formula:** $$A = L \times \left(\frac{P}{2} - L\right) = \frac{P}{2}L - L^2$$ 5. **Maximize the area:** This is a quadratic function in $L$: $$A(L) = -L^2 + \frac{P}{2}L$$ The maximum value of a quadratic $ax^2 + bx + c$ with $a<0$ is at $x = -\frac{b}{2a}$. Here, $a = -1$, $b = \frac{P}{2}$, so $$L = -\frac{\frac{P}{2}}{2 \times (-1)} = \frac{P}{4}$$ 6. **Find $W$:** $$W = \frac{P}{2} - L = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$$ 7. **Conclusion:** The dimensions that maximize the area are: $$L = W = \frac{P}{4}$$ This means the enclosure is a square. **Final answer:** The rectangular enclosure with maximum area for a given perimeter is a square with side length $\frac{P}{4}$.