13. A rectangular playing field is to have an area of 600 m². Fencing is required to enclose the field and divide it into two equal halves.

a) Find a formula, [tex]\( F(x) \)[/tex], for the total length of fencing required, in terms of the length [tex]\( x \)[/tex] of the fence dividing the field in half.

b) Find the minimum amount of fencing needed to do this.

c) What are the outer dimensions of the field that require the least amount of fencing?



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]