How Can a Farmer Maximize the Enclosed Area with 120 Meters of Fencing Next to a River?

Question

A farmer has 120 meters of fencing and wants to enclose a rectangular field next to a straight river, with the river acting as one side of the boundary. To minimize fencing costs and maximize the use of the available fencing, the farmer plans to use the river as one of the sides of the enclosure. The goal is to determine the dimensions that will maximize the enclosed area, using the fencing only for the other three sides (two shorter sides and the opposite longer side). What dimensions will give the largest enclosed area? Additionally, what is the maximum area that can be achieved?

Key Considerations:

1. The river provides a natural boundary, so no fencing is needed for one side of the rectangle.
2. The farmer has a fixed amount of fencing (120 meters) for the remaining three sides.
3. The task is to find the optimal dimensions (length and width) that maximize the area of the enclosure.

By solving this problem, we can determine not only the ideal dimensions but also the maximum area that can be enclosed with the available fencing.

That's Not My Neighbor

Answer 1

Let x represent the two sides that are perpendicular to the river. Let y represent the length of the side parallel to the river. The amount of fencing is then 2x + y. Since we know that the fencing available is 120 meters, we have this constraint on x and y:

2x + y = 120

Solve that equation for y to get y in terms of x:

y = 120 - 2x

The area of the rectangle is A=LW or in this case A = xy. Since we have an expression for y in terms of x we can represent the area depending on x only:

A = x(120 - 2x)

A = -2x² + 120x

This is a quadratic function so its graph is a parabola. Since its leading coefficient (-2) is negative the parabola opens downward. That means the coordinates of its vertex give the required information. To find the coordinates of the vertex there are a few methods available. One method is to realize that the x coordinate of the vertex is located along its axis of symmetry and that the equation of the axis of symmetry is given by x = -b/2a where a = (-2) and b = (120).

x = -120/2(-2) = -120/-4 = 30

When x = 30, y = 120 - 2(30) = 60 meters

A(30) = -2(30²) + 120(30) = -1800 + 3600 =1800 m²

Note that is you solve for the roots of the quadratic function:

0 = x(120-2x)

x = 0 or x = 60, the vertex lies along the line x = c where c is the average of the roots, i.e. (0+60)/2 = 30. Just another way to determine the x coordinate of the vertex.

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