A rancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms

Let x be the shorter side, and y be the longer side

There would be 4 fences along the shorter side, and 2 fences along the longer side

4x + 2y = 800

Rewrite in terms of y:

y = 400 − 2x

The area of the rectangular field is

A = x*y

Replace Y with the equation above:

  A = x(400 − 2x)

  A = − 2x^2 + 400x

The area is a parabola that opens downward, the maximum area would occur at the parabola vertex.

At the vertex

x = −b/2a

  = −400/[2(−2)]

  = 100

y = 400 −2x

y = 400 -2(100)

y = 400-200

y = 200

The dimension of the rectangular field that maximize the enclosed area is 100 ft x 200 ft.

The dimension of the rectangular field that maximize the enclosed area is 100 ft by 200 ft.

Application of the vertex of a quadratic equation

The formula for calculating the vertex of  a quadratic equation is given as:

x = -b/2a

From the given question

Let the shorter side be x

Let the longer side of the field be y

If there are 4 fences along the shorter side, and 2 fences along the longer side, hence;

4x + 2y = 800

Write the resultinq equation in slope-intecept form

2y = -4x + 800

y = -2x + 400

y = 400 − 2x

Determine the area of the field

Area = x*y

Substitute the expression above into the area to have:

A = x(400 − 2x)

A = − 2x^2 + 400x

Since the parabola opens downward, the maximum area would occur at the parabola vertex as shown;

x = −b/2a

x = −400/2(-2)

x = 400/4

x = 100

Determine the value of y

y = 400 −2x

y = 400 -2(100)

y = 400-200

y = 200

Hence the dimension of the rectangular field that maximize the enclosed area is 100 ft by 200 ft.

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