Answer :
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]
