1. **Problem statement:** A farmer has 450 m of fencing to enclose a rectangular area and divide it into two sections with a fence parallel to one side. 2. **Define variables:** Let the width be $w$ meters and the length be $l$ meters. 3. **Fencing constraint:** The fencing encloses the rectangle and divides it into two sections, so the total fencing used is for three widths and two lengths: $$3w + 2l = 450$$ 4. **Express length in terms of width:** $$2l = 450 - 3w \implies l = \frac{450 - 3w}{2}$$ 5. **Area function:** The total area $A$ is length times width: $$A(w) = l \times w = w \times \frac{450 - 3w}{2} = \frac{450w - 3w^2}{2} = 225w - \frac{3}{2}w^2$$ 6. **Domain:** Width $w$ must be positive and length $l$ must be positive: - From $w > 0$ - From $l > 0 \Rightarrow \frac{450 - 3w}{2} > 0 \Rightarrow 450 - 3w > 0 \Rightarrow w < 150$ So domain is: $$0 < w < 150$$ 7. **Range:** The area function is a quadratic opening downward. The maximum area occurs at the vertex. 8. **Find vertex:** For $$A(w) = -\frac{3}{2}w^2 + 225w$$ The vertex $w$ coordinate is: $$w = -\frac{b}{2a} = -\frac{225}{2 \times (-\frac{3}{2})} = \frac{225}{3} = 75$$ 9. **Maximum area:** Substitute $w=75$: $$A(75) = 225 \times 75 - \frac{3}{2} \times 75^2 = 16875 - \frac{3}{2} \times 5625 = 16875 - 8437.5 = 8437.5$$ 10. **Corresponding length:** $$l = \frac{450 - 3 \times 75}{2} = \frac{450 - 225}{2} = \frac{225}{2} = 112.5$$ 11. **Range of $A$:** Since $A(0) = 0$ and $A(150) = 0$, and maximum is $8437.5$, the range is: $$0 < A(w) \leq 8437.5$$