Thinkin problem

Question

Farmer Ed has 300 meters of fencing and wants to enclose a rectangular plot that borders a river. If Farmer Ed does not fence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?

Answer 1

parallel side = 150 meters

perpendicular side = 75 meters

max area = 150(75) = 11,250 m^2

Area = Lw L+2w =300, L = 300-2w

Area = (300-2w)w =300w -2w^2

take the derivative and set = 0

A' = 300-4w = 0

4w =300

w = 300/4 = 75 meters wide

L = 300-2w = 300- 2(75) = 300-150 = 150 meters Long

max Area = 75x150 = 11,250 m^2

Answer 2

There have been several problems posted similar to this one. Here is one way to solve without using Calculus.

Let x represent the length of the two sides that are perpendicular to the river.

Let y represent the length of the side that is parallel to the river.

Then the total length of fencing is represented by 2x + y. We know there are 300 feet of fencing available.

2x + y = 300

y = 300 - 2x

The area of the rectangular plot is length times width: A = xy. By substituting 300 - 2x for y we have the area dependent on one variable (x).

A = x(300 - 2x)

The graph of this function is a parabola that opens downward. Locating the vertex of that parabola will help answer the questions. Note the y coordinate of the vertex gives the maximum value for A. There are several ways to locate the vertex.

Since the function definition is already in factored form, one way is to determine its roots. The axis of symmetry passes through the midpoint of the segment joining the roots along the x-axis.

x(300-2x) = 0

x = 0 or 300 - 2x = 0

x = 0 or x = 150

The midpoint is x = (150+0)/2 = 75. The equation of the axis of symmetry is also x = 75. The vertex lies on that line.

A = (75)[300 - 2(75)] = (75)(150) = 11250 ft²

So when x = 75, A = 11250 (a maximum).

The y dimension is given by y = 300 - 2x = 150.

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