a rancher has 1200 fet of fencing to enclose two adjacent cattle pens. What dimensions should be used so that the enclosed area is maximized?
You don't describe the diagram, but I am familiar with this type of question.
Assuming there is a common width between the two fields
let the width be x ft
let the length of the combined field be y
so
3x + 2y = 1200
y = 600 - (3/2)x
area = xy
= x(600 - (3/2)x)
= 600x - (3/2)x^2
d(area)/dx = 600 - 3x
= 0 for a max of area
3x = 600
x = 200
y = 600-(3/2)(200) = 300
state the conclusion
To maximize the enclosed area, the rancher should divide the fencing into two equal parts since the two cattle pens are adjacent. Let's call the length of each adjacent side x.
Since the rancher has a total of 1200 feet of fencing, the total length of all four sides will be 1200 feet. Since there are two adjacent sides on each pen and they have equal length, the equation for the perimeter can be written as:
2x + 2x = 1200
Simplifying the equation further, we get:
4x = 1200
Dividing both sides of the equation by 4, we find:
x = 300
Therefore, each adjacent side should be 300 feet long in order to maximize the enclosed area.
To find the dimensions that would maximize the enclosed area, we can use the concept of optimization. Let's break down the problem into smaller steps:
1. Define the variables: Let's assume the width of one pen is "w" (in feet). Since there are two adjacent pens, the total width of both pens will be "2w." The length of both pens combined will be "l," which we need to find.
2. Formulate the equation for the area: The area of one pen can be calculated as length multiplied by width. Since we have two pens, the total area becomes: A = w * l * 2.
3. Set up the constraint: The total amount of fencing available is 1200 feet. We can establish a constraint based on this: 2w + l + l = 1200, which can be simplified to 2w + 2l = 1200, or w + l = 600.
4. Solve the constraint for one variable: We can solve the constraint equation for one variable and substitute it into the area equation. It's easier here to solve for "w" in terms of "l": w = 600 - l.
5. Substitute the expression for "w" into the area equation: A = (600 - l) * l * 2.
6. Find the maximum value of the area: To find the maximum area, we can differentiate the area equation with respect to "l" and set it equal to zero. Then we solve for "l."
Taking the derivative of the area equation:
dA/dl = (600 - l) * 2 - 2l
0 = 1200 - 4l
4l = 1200
l = 300
7. Calculate the width: Substituting the value of "l" back into the constraint equation gives us the value of "w": w = 600 - 300 = 300.
Hence, the dimensions that will maximize the enclosed area are a width of 300 feet and a length of 300 feet for each pen.