A cattle rancher wants to enclose a rectangular area and then divide it into six pens with fencing parallel to one side of the rectangle (see the figure below). There are 480 feet of fencing available to complete the job. What is the largest possible total area of the six pens?
finally I get to use calculus for this problem instead of completing square!
length total = 6 x
width = y
fencing = 7 y + 12 x = 480
so
y = (480-12x)/7
A = 6 x y
A = 6 (x)(480-12 x)/7
A = (6/7) (480 x - 12 x^2 )
dA/dx = (6/7)(480 - 24 x) = for max
24 x = 480
x = 20
then y = 240/7 = 34.3 approx
A = 6 x y = 6 * 20 * 240/7 = 4114.3
I have done many of these fencing problems tonight but most were in algebra where I had to find the vertex of the parabola by completing the square instead of taking the derivative and setting it to zero. Using Calculus makes it much much easier.
Why did the cattle rancher want to enclose the rectangular area and divide it into six pens? Because he wanted to create a "ranchy" neighborhood for his cows to socialize and have parties! Now, let's calculate the largest possible total area of the six pens.
Let's say the length of the rectangular area is L and the width is W. Since there are six pens, we'll divide the rectangular area into three equal parts.
To maximize the area, we should make the width of the rectangular area as small as possible, which means the three pens on the short side of the rectangle will each have a width of W/3.
The remaining three pens will be located on the long side of the rectangle, and each will have a length of L/2 since they are adjacent to each other.
Now let's calculate the length of fencing required:
- The long side of the rectangular area: L + 2L/2 = 2L
- The short side of the rectangular area: 2(W/3) + 2(W/3) + 2(W/3) = 2W
Since we have a total of 480 feet of fencing available, we can write the equation:
2L + 2W = 480
Now, let's express L in terms of W:
2L = 480 - 2W
L = (480 - 2W)/2
L = 240 - W
We want to find the largest possible total area, which is A = 3(L/2)(W/3) + 3(L/2)(W/3) + (L)(W/3) = LW/2 + LW/2 + LW/3 = (5LW)/6
Substituting the value of L from the equation above, we get:
A = [(5(240 - W))/6]W
To find the maximum area, we take the derivative of A with respect to W and set it equal to zero:
dA/dW = (1200 - 5W)/6 = 0
Solving for W, we get W = 1200/5 = 240
Substituting the value of W back into the equation for L:
L = 240 - W = 240 - 240 = 0
Oh no! It seems like L is zero, which means we can't have a rectangular area. Maybe the cattle rancher should consider organizing a comedy show instead of creating pens. The cows would appreciate some laughter while waiting to be milked! 🤡
Let's solve the problem step-by-step:
Step 1: Determine the length of the rectangle.
Let's assume the length of the rectangle is "L" in feet.
Step 2: Determine the width of the rectangle.
Let's assume the width of the rectangle is "W" in feet.
Step 3: Calculate the perimeter of the rectangle.
The perimeter of the rectangle is given by:
Perimeter = 2L + W
Since there are six pens and the fencing is parallel to one side of the rectangle, the perimeter of the rectangle is equal to the sum of the lengths of the six pens:
Perimeter = 6L
Step 4: Write the equation for the perimeter of the rectangle.
From step 3, we know that the perimeter of the rectangle is 6L. And from the problem statement, we know that the total available fencing is 480 feet. Therefore, we can write the equation:
6L = 480
Step 5: Solve the equation for L.
Dividing both sides of the equation by 6:
L = 480/6
L = 80
Step 6: Calculate the width of the rectangle.
Since there are six pens, the width of the rectangle is equal to the length of one pen. Therefore, the width is:
W = L/6
W = 80/6
W = 13.33 (rounded to two decimal places)
Step 7: Calculate the area of one pen.
The area of one pen is given by:
Area = Length x Width
Area = L x W
Area = 80 x 13.33
Area = 1066.4 (rounded to one decimal place)
Step 8: Calculate the total area of the six pens.
Since there are six pens, the total area is given by:
Total Area = Area of one pen x Number of pens
Total Area = 1066.4 x 6
Total Area = 6398.4 (rounded to one decimal place)
Therefore, the largest possible total area of the six pens using the available fencing of 480 feet is approximately 6398.4 square feet.
To find the largest possible total area of the six pens, we need to optimize the dimensions of the rectangular area. Let's break down the problem step by step:
1. Assume the length of the rectangle is x feet.
2. Since the rectangular area is divided into six pens with fencing parallel to one side, we can conclude that the width of the rectangle is x/6 feet.
3. The perimeter of the rectangular area can be calculated by adding up the lengths of all four sides:
Perimeter = 2*(length + width) = 2*(x + x/6) = 2*(7x/6) = 7x/3 feet.
4. According to the problem, the rancher has 480 feet of fencing available, so we can equate the perimeter to 480:
7x/3 = 480.
To solve for x, we can multiply both sides by 3/7:
x = (480*3)/7 = 206.86 feet (approximately).
So, the length of the rectangle is approximately 206.86 feet.
Now, to find the width of the rectangle, we divide the length by 6:
Width = 206.86/6 = 34.48 feet (approximately).
To find the area of one pen, we multiply the length and width:
Area of one pen = length * width = 206.86 * 34.48 = 7133.07 square feet (approximately).
Since we have six pens, the total area of all six pens is:
Total area = Area of one pen * 6 = 7133.07 * 6 = 42,798.42 square feet (approximately).
Therefore, the largest possible total area of the six pens when 480 feet of fencing is available is approximately 42,798.42 square feet.