Questions: A city decides to make a park by fencing off a section of riverfront property. Funds are allotted to provide 80 meters of fence. The area enclosed will be a rectangle, but only three sides will be enclosed by fence - the other side will be bound by the river. What is the maximum area that can be enclosed in this way? a) Make a rough sketch of the park b) Decide how you will represent the dimensions. c) Develop a polynomial function to describe the area to be enclosed. d) Solve the problem showing all algebra work and include a graph of the polynomial with axes labeled and calibrated and key points shown. Please remember to title your graph.

Solution Steps

Let $x$ be the length of the side parallel to the river, and $y$ be the length of each of the two sides perpendicular to the river.

The total length of the fence is given by: $$x + 2y = 80$$ Solving for $x$: $$x = 80 - 2y$$ The area $A$ of the rectangle is: $$A = x \cdot y = (80 - 2y) \cdot y = 80y - 2y^2$$

To find the maximum area, we need to find the vertex of the parabola described by the quadratic function $A(y) = 80y - 2y^2$. The vertex form of a parabola $ay^2 + by + c$ has its maximum (or minimum) at $y = -\frac{b}{2a}$.

Here, $a = -2$ and $b = 80$: $$y = -\frac{80}{2(-2)} = \frac{80}{4} = 20$$

Substituting $y = 20$ back into the equation for $x$: $$x = 80 - 2(20) = 40$$

The maximum area is: $$A = 40 \cdot 20 = 800$$

Final Answer