(Six pigpen problem). A 2 × 3 array of six congruent rectangular pigpens (that all look the same from above) will be in the overall shape of a rectangle R. We may use 100 feet of fencing to form the boundaries of the pigpens. Find the dimensions for a single pigpen that will maximize the total area of all the pigpens, and find this total area. (The fencing separating the pigpens has constant height, so we may ignore height in our calculations. Also, assume the boundaries between pigpens are not double - fenced; that is, assume that the thickness of the fencing between pigpens is the same as the thickness of the fencing along the outer boundary, R .)
My numbers are not coming at the same as my professor. I have redid this problem multiple times and I just cannot get it. Please explain.
show what you did. This kind of problem always works out that the fencing is divided equally among lengths and widths. So, the entire array will have dimension
(50/3) by (50/4)
Each pen will be (50/9) by (50/8)
Start with
3x+4y = 100
a = xy = x*(100-3x)/4
then find x such that da/dx=0
To solve this problem, we need to find the dimensions of a single pigpen that will maximize the total area of all the pigpens.
Let's start by considering the perimeter of a single pigpen. Since all pigpens are congruent and the overall shape is a rectangle, we can write the equation:
2l + 3w = 100
where l represents the length of the pigpen and w represents the width. This equation represents the total length of the fencing needed to enclose the pigpen. We have a total of 100 feet of fencing available.
To maximize the total area of all the pigpens, we need to optimize the dimensions of a single pigpen. The area of a rectangle is given by the equation:
A = l * w
Since we want to maximize this area, we need to express it in terms of a single variable. For this, we can use the equation for the perimeter:
l = (100 - 3w)/2
Substituting this into the area equation, we get:
A = (100 - 3w)/2 * w
To find the dimensions that maximize the total area, we can take the derivative of this expression with respect to w and set it equal to zero. This will give us a critical point where the maximum occurs.
dA/dw = (100 - 6w)/2
Setting this equal to zero and solving for w, we get:
100 - 6w = 0
6w = 100
w = 100/6 ≈ 16.67
Now that we have the width of a single pigpen, we can substitute this value back into the equation for the perimeter to find the length:
l = (100 - 3w)/2
l = (100 - 3*16.67)/2 ≈ 33.33
Therefore, the dimensions for a single pigpen that will maximize the total area of all the pigpens are approximately 16.67 feet for the width and 33.33 feet for the length.
To find the total area of all the pigpens, we multiply the area of a single pigpen by the total number of pigpens. In this case, we have a 2x3 array, so the total number of pigpens is 2 * 3 = 6.
Total area = Area of single pigpen * total number of pigpens
Total area = (16.67 * 33.33) * 6 ≈ 3333.33 square feet