A rancher has 5,400 linear feet of fencing and wants to enclose a rectangular field and then divide it into two equal pastures with three internal fences parallel to one of the rectangular sides. What is the maximum area of each pasture?
The Correct Answer and Explanation is :
Solution:
The rancher has 5,400 linear feet of fencing and wants to maximize the area of two equal pastures. Let:
- ( L ): Length of the rectangle
- ( W ): Width of the rectangle (parallel to the internal dividing fences)
Step 1: Total fencing usage
The rancher needs fencing for:
- Two lengths (( 2L )): the two longer sides of the rectangle.
- Four widths (( 4W )): two outer widths plus three internal dividing fences.
The total fencing is:
[
2L + 4W = 5400 \quad \text{(Equation 1)}
]
Step 2: Solve for ( L )
From Equation 1:
[
L = \frac{5400 – 4W}{2} = 2700 – 2W \quad \text{(Equation 2)}
]
Step 3: Area of the rectangle
The area of the rectangle is:
[
A = L \cdot W
]
Substitute ( L = 2700 – 2W ) into the area equation:
[
A = (2700 – 2W) \cdot W
]
[
A = 2700W – 2W^2
]
Step 4: Maximize the area
To maximize ( A ), take the derivative and set it equal to zero:
[
\frac{dA}{dW} = 2700 – 4W = 0
]
[
W = 675
]
Substitute ( W = 675 ) into Equation 2 to find ( L ):
[
L = 2700 – 2(675) = 1350
]
Step 5: Total area and area per pasture
The total area of the rectangle is:
[
A = L \cdot W = 1350 \cdot 675 = 911,250 \, \text{square feet}
]
Each pasture has half this area:
[
\text{Area per pasture} = \frac{911,250}{2} = 455,625 \, \text{square feet}
]
Explanation (300 Words)
The rancher has a fixed amount of fencing, 5,400 feet, and seeks to maximize the enclosed area. Since the field is divided into two pastures by three internal fences, the fencing formula includes the lengths of the rectangle (( 2L )) and four widths (( 4W )). This results in the constraint equation ( 2L + 4W = 5400 ), which ensures all fencing is used.
The area ( A ) of the rectangle depends on the length ( L ) and width ( W ). Substituting ( L = 2700 – 2W ) (derived from the fencing constraint) into the area formula ( A = L \cdot W ), we express ( A ) as a quadratic function: ( A = 2700W – 2W^2 ). This quadratic has a maximum value because the coefficient of ( W^2 ) is negative, forming a parabola that opens downward.
The maximum area occurs at the vertex of the parabola, found using calculus (( \frac{dA}{dW} = 0 )) or the vertex formula. Solving for ( W ), we find ( W = 675 ), and substituting back, ( L = 1350 ). The total area is 911,250 square feet, and since the field is split into two equal pastures, each pasture’s area is 455,625 square feet.
This solution demonstrates optimal resource use while balancing geometry and constraints.