1. Farmer Brown has 1200ft of fence to create a rectangular pen.
If he would like to separate the pen into 4 areas, what dimensions will maximize and what is the maximum area? (Do not forget units!)
Let the pen have sides length a and b (both in ft)
If we assume that the sides for the separate internal pens are parallel to the sides of the outer pen then
3a+3b=1200 ft or a+b=400 ft
and the area=ab or a=area/b
substitute in the equation above
area/b + b = 400 ft
area = 400b-b^2
which is a maximum when
400 ft = 2b or b=200 ft, hence a=200 ft
But check the maths!
To find the dimensions that will maximize the area of the rectangular pen and calculate the maximum area, you can utilize the derivative and optimization principles. Here's how you can do it step by step:
Step 1: Understand the problem.
Farmer Brown has 1200ft of fence to create a rectangular pen. The pen needs to be divided into four equal areas. We need to find the dimensions of the pen that will maximize the area, and calculate the maximum area.
Step 2: Define the problem algebraically.
Let the length of the pen be denoted by 'L' and the width be denoted by 'W'. We need to find the dimensions L and W that maximize the area of the rectangular pen (A) while using 1200ft of fencing.
Step 3: Set up the equations.
We know that the perimeter of a rectangle is given by: Perimeter = 2L + 2W.
Therefore, from the given problem, we can set up the equation:
2L + 2W = 1200 (Equation 1)
Since the pen needs to be divided into four equal areas, we can find the dimensions of just one of those four areas. In that case, the area of the rectangular pen (A) is given by:
A = L * W.
Step 4: Solve for one variable in terms of the other.
Using Equation 1, we can solve for one variable in terms of the other. Let's solve for L:
2L = 1200 - 2W
L = (1200 - 2W)/2
L = 600 - W
Step 5: Rewrite the area function.
Using the equation for L from Step 4, we can rewrite the area function in terms of W:
A = (600 - W) * W
A = 600W - W^2
Step 6: Maximize the area.
To maximize the area, we need to find the value of W that maximizes the area function A. We can do this by taking the derivative of A with respect to W, setting it equal to zero, and solving for W.
dA/dW = 600 - 2W (Differentiate A with respect to W)
600 - 2W = 0 (Set dA/dW equal to zero to find critical points)
-2W = -600
W = 300
Step 7: Calculate the dimensions and maximum area.
Now that we have the value of W, we can substitute it back into Equation 1 to find the value of L:
2L + 2W = 1200
2L + 2(300) = 1200
2L + 600 = 1200
2L = 600
L = 300
Therefore, the value of W is 300ft and the value of L is also 300ft. The maximum area (A) is given by:
A = L * W = 300 * 300 = 90000 square feet.
So, the dimensions that maximize the area are a length of 300ft and a width of 300ft, and the maximum area is 90000 square feet.