A rectangular study area is to be enclosed by a fence and divided into two equal parts, with a fence running along the division parallel to one of the sides. if the total area is 384 ft^2, find the dimensions of the study area that will minimize the total length of the fence. How much fence will be required?
p = perimeter = 3 L + 2 B
384 = L B so B = 384/L
p = 3 L + 768/L
dp/dL = 3 - 786 /L^2
= 0 for minimum
3 L^2 = 786
L^2 = 256
L = 16
B = 384/16 = 24
B/2 = 12
so two 12 by 16 areas
To find the dimensions of the study area that will minimize the total length of the fence, we can use optimization techniques by differentiating and finding the critical points.
Let's assume the width of the study area is "x" ft and the length is "y" ft.
It is given that the study area is rectangular and divided into two equal parts by a fence running parallel to one of the sides. This means one part will have dimensions x ft by y/2 ft, and the other part will also have dimensions x ft by y/2 ft.
The total area of the study area is given as 384 ft², so we have the equation:
x * y = 384
Now, we need to express the length of the fence in terms of x and y. The fence will run along the width of the study area and also along the division parallel to the length.
The length of the fence along the width is 2x ft since there are two equal parts.
The length of the fence along the division parallel to the length is y ft since the division runs along the y-axis.
Therefore, the total length of the fence, L, can be expressed as:
L = 2x + y
To minimize the total length of the fence, we need to find the critical points by differentiating L with respect to x and y, and setting the partial derivatives equal to zero.
Differentiating L with respect to x:
dL/dx = 2
Differentiating L with respect to y:
dL/dy = 1
Setting dL/dx = 0 and dL/dy = 0, we get:
2 = 0 => This is not possible.
1 = 0 => This is not possible.
Since no critical points exist, the minimum length of the fence occurs at the boundaries or endpoints.
From the given equation, x * y = 384, we can solve for y in terms of x:
y = 384 / x
Now, substituting this value of y into the equation for the total length of the fence, we have:
L = 2x + (384 / x)
To find the value of x that minimizes L, we can find the derivative of L with respect to x and set it equal to zero:
dL/dx = 2 - 384/x² = 0
Solving this equation, we get:
2 - 384/x² = 0
384/x² = 2
x² = 384/2
x² = 192
x = √192
x ≈ 13.86 ft
Substituting this value of x into the equation for y, we have:
y = 384 / x
y = 384 / 13.86
y ≈ 27.69 ft
Therefore, the dimensions of the study area that will minimize the total length of the fence are approximately 13.86 ft by 27.69 ft.
To find the total length of the fence, substitute these values in the equation for L:
L = 2x + y
L = 2(13.86) + 27.69
L ≈ 55.41 ft
So, approximately 55.41 ft of fence will be required.