Max Area Rectangle 1Ce695

1. **State the problem:** Sam has 120 feet of fencing to build a rectangular pen. We want to find the length and width that maximize the area of the rectangle. 2. **Formula and rules:** - Let the length be $L$ and the width be $W$. - The perimeter $P$ of a rectangle is given by $$P = 2L + 2W$$ - The area $A$ of a rectangle is $$A = L \times W$$ - We know the perimeter is fixed: $$2L + 2W = 120$$ 3. **Express one variable in terms of the other:** $$2L + 2W = 120 \implies L + W = 60 \implies W = 60 - L$$ 4. **Write area as a function of one variable:** $$A = L \times W = L(60 - L) = 60L - L^2$$ 5. **Maximize the area function:** - The area function is $$A(L) = -L^2 + 60L$$ which is a quadratic opening downward. - The vertex of a quadratic $$ax^2 + bx + c$$ is at $$L = -\frac{b}{2a}$$. - Here, $$a = -1$$ and $$b = 60$$, so $$L = -\frac{60}{2 \times (-1)} = \frac{60}{2} = 30$$ 6. **Find corresponding width:** $$W = 60 - L = 60 - 30 = 30$$ 7. **Conclusion:** - The maximum area is achieved when length and width are both 30 feet. - This means the pen is a square. **Final answer:** $$\boxed{L = 30 \text{ feet}, \quad W = 30 \text{ feet}}$$