a total of 8 meters of fencing are going to be used to fence in a rectangular cage for pets and divide it into three smaller cages
**the photo that is shown with this question is three (3) boxes stuck together side by side horizontally***
determine the overall dimensions that will yield the maximum total enclosed area.
Let p=Length; n =width
8=2n+4p;
p=2-n/2
Area=np =n(2-n/2)=2n-(n^2)/2
Derivative:2-n
Critical value: 2-n=0
n=2
P=2-2/2=1
1meter by 2 meter
To determine the overall dimensions that will yield the maximum total enclosed area, we need to use the concept of optimization. In this case, we want to maximize the area while using a fixed amount of fencing (8 meters).
Let's suppose the larger rectangular cage has length L and width W. Based on the information provided, we can deduce the following:
1. The three smaller cages will be created by dividing the larger rectangular cage into equal sections horizontally.
2. The length of each smaller cage will be L, while the width will be W/3.
Now, let's express the total amount of fencing used in terms of the dimensions, L and W:
Perimeter of the larger rectangular cage = 2L + 2W
We know that the total amount of fencing available is 8 meters, so we have:
2L + 2W = 8
L + W = 4 (equation 1)
Next, let's calculate the area of the larger rectangular cage in terms of L and W:
Area of the larger rectangular cage = L * W
Since we have three smaller cages of equal length (L) and width (W/3), the total enclosed area will be three times the area of a smaller cage:
Total enclosed area = 3 * (L * (W/3)) = L * W (equation 2)
To find the dimensions that maximize the overall enclosed area, we need to optimize equation 2 with respect to equation 1.
We can re-arrange equation 1 to express L in terms of W:
L = 4 - W
Substituting this expression for L in equation 2, we get:
Total enclosed area = (4 - W) * W
To find the maximum area, we need to find the value of W that maximizes this equation. We can do this by taking the derivative of the equation with respect to W, setting it equal to zero, and solving for W.
d(Area)/dW = 0
4 - 2W = 0
2W = 4
W = 2
Now that we have the value of W, we can substitute it back into equation 1 to find the corresponding value of L:
L + 2 = 4
L = 2
Therefore, the overall dimensions that will yield the maximum total enclosed area with 8 meters of fencing are:
Length (L) = 2 meters
Width (W) = 2 meters
Thus, the larger rectangular cage will be 2 meters long and 2 meters wide, while each smaller cage will be 2 meters long and (2/3) meters wide.