Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Certainly! Let's tackle each part of this question step by step.
### Part A: Finding the Formula for the Total Length of Fencing Required
Given:
- The area of the rectangular field is [tex]\( 600 \, \text{m}^2 \)[/tex].
- The field is to be divided into two equal halves by a fence.
Let's denote:
- The length dividing the field in half as [tex]\( x \)[/tex].
- The width of the field as [tex]\( y \)[/tex].
Since the area of the field is 600 [tex]\( \text{m}^2 \)[/tex]:
[tex]\[ x \times y = 600 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{600}{x} \][/tex]
The total length of fencing required includes:
1. The two lengths [tex]\( y \)[/tex] (each side of the divided halves).
2. The two widths [tex]\( x \)[/tex] (the top and bottom of the rectangle).
3. One additional length [tex]\( y \)[/tex] (the dividing fence).
So, the total length of fencing [tex]\( F(x) \)[/tex] is:
[tex]\[ F(x) = 2y + 2x + y \][/tex]
[tex]\[ F(x) = 3y + 2x \][/tex]
Substituting [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ F(x) = 3 \left(\frac{600}{x}\right) + 2x \][/tex]
Thus, the formula for the total length of fencing required is:
[tex]\[ F(x) = \frac{1800}{x} + 2x \][/tex]
### Part B: Finding the Minimum Amount of Fencing Needed
To find the minimum amount of fencing needed, we need to minimize the function [tex]\( F(x) = \frac{1800}{x} + 2x \)[/tex].
Through calculus or numerical optimization methods, it is found that the minimum value of [tex]\( F(x) \)[/tex] is:
[tex]\[ \min F(x) = 120 \, \text{m} \][/tex]
### Part C: Finding the Outer Dimensions of the Field with the Least Fencing
To find the corresponding values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that minimize the function:
Given our minimization result:
[tex]\[ x = 30 \, \text{m} \][/tex]
[tex]\[ y = 20 \, \text{m} \][/tex]
So, the outer dimensions of the field that require the least amount of fencing are:
[tex]\[ 20 \, \text{m} \times 30 \, \text{m} \][/tex]
### Part A: Finding the Formula for the Total Length of Fencing Required
Given:
- The area of the rectangular field is [tex]\( 600 \, \text{m}^2 \)[/tex].
- The field is to be divided into two equal halves by a fence.
Let's denote:
- The length dividing the field in half as [tex]\( x \)[/tex].
- The width of the field as [tex]\( y \)[/tex].
Since the area of the field is 600 [tex]\( \text{m}^2 \)[/tex]:
[tex]\[ x \times y = 600 \][/tex]
Solving for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{600}{x} \][/tex]
The total length of fencing required includes:
1. The two lengths [tex]\( y \)[/tex] (each side of the divided halves).
2. The two widths [tex]\( x \)[/tex] (the top and bottom of the rectangle).
3. One additional length [tex]\( y \)[/tex] (the dividing fence).
So, the total length of fencing [tex]\( F(x) \)[/tex] is:
[tex]\[ F(x) = 2y + 2x + y \][/tex]
[tex]\[ F(x) = 3y + 2x \][/tex]
Substituting [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ F(x) = 3 \left(\frac{600}{x}\right) + 2x \][/tex]
Thus, the formula for the total length of fencing required is:
[tex]\[ F(x) = \frac{1800}{x} + 2x \][/tex]
### Part B: Finding the Minimum Amount of Fencing Needed
To find the minimum amount of fencing needed, we need to minimize the function [tex]\( F(x) = \frac{1800}{x} + 2x \)[/tex].
Through calculus or numerical optimization methods, it is found that the minimum value of [tex]\( F(x) \)[/tex] is:
[tex]\[ \min F(x) = 120 \, \text{m} \][/tex]
### Part C: Finding the Outer Dimensions of the Field with the Least Fencing
To find the corresponding values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that minimize the function:
Given our minimization result:
[tex]\[ x = 30 \, \text{m} \][/tex]
[tex]\[ y = 20 \, \text{m} \][/tex]
So, the outer dimensions of the field that require the least amount of fencing are:
[tex]\[ 20 \, \text{m} \times 30 \, \text{m} \][/tex]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
