1. Problem 1: Determine the equation modeling the steel arches spanning a river 281 m wide with a height of 71 m. 2. The arch can be modeled as a parabola symmetric about the midpoint of the river. 3. Let the horizontal axis $x$ be centered at the midpoint of the river, so $x=0$ is the center. 4. The arch touches the river at $x=\pm \frac{281}{2} = \pm 140.5$ m, where the height $y=0$. 5. The maximum height is $y=71$ m at $x=0$. 6. The parabola equation form is $y = a x^2 + c$. 7. Since the vertex is at $(0,71)$, $c=71$. 8. Using the point $(140.5,0)$, substitute into the equation: $$0 = a (140.5)^2 + 71$$ 9. Solve for $a$: $$a = -\frac{71}{(140.5)^2} = -\frac{71}{19740.25} \approx -0.0036$$ 10. Therefore, the equation modeling the arch is: $$y = -0.0036 x^2 + 71$$ 11. Units: $x$ in meters (horizontal distance from center), $y$ in meters (height above river). 12. Therefore, the steel arch can be modeled by the parabola $$y = -0.0036 x^2 + 71$$ where $x$ ranges from $-140.5$ m to $140.5$ m. 13. Problem 2: Find the dimensions of a rectangular fence with perimeter 120 m that maximize the enclosed area. 14. Let the length be $L$ meters and the width be $W$ meters. 15. The perimeter constraint is: $$2L + 2W = 120 \implies L + W = 60$$ 16. Express $L$ in terms of $W$: $$L = 60 - W$$ 17. The area $A$ is: $$A = L \times W = (60 - W) W = 60W - W^2$$ 18. To maximize area, find the vertex of the parabola $A(W) = -W^2 + 60W$. 19. The vertex $W$ coordinate is: $$W = -\frac{b}{2a} = -\frac{60}{2(-1)} = 30$$ 20. Substitute $W=30$ back to find $L$: $$L = 60 - 30 = 30$$ 21. Therefore, the dimensions that maximize area are $L=30$ m and $W=30$ m. 22. The maximum area is: $$A = 30 \times 30 = 900 \text{ m}^2$$ 23. Therefore, the fence should be a square 30 m by 30 m to maximize the enclosed area with 120 m of fencing.