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### 3.1. Length of the Side of the Square:
Consider a piece of wire that is 10 meters long. Let the length of the wire used to make the square be [tex]\(x\)[/tex] meters.
To form a square, the wire is bent into 4 equal parts. Therefore, the length of one side of the square ([tex]\(s\)[/tex]) in terms of [tex]\(x\)[/tex] is:
[tex]\[ s = \frac{x}{4} \][/tex]
### 3.2. Sum of the Areas of the Square and the Rectangle:
Next, we need to show that the sum of the areas of the square and the rectangle is given by [tex]\(S(x) = -\frac{1}{8} x^2 + \frac{5}{4} x\)[/tex].
#### Area of the Square:
The area of the square ([tex]\(A_{\text{square}}\)[/tex]) is:
[tex]\[ A_{\text{square}} = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \][/tex]
#### Area of the Rectangle:
The remaining length of the wire, that is not used for the square, is used for the rectangle. Therefore, the length of the wire used for the rectangle is:
[tex]\[ 10 - x \text{ meters} \][/tex]
Given that the width of the rectangle ([tex]\(w\)[/tex]) is the same as the side length of the square, we have:
[tex]\[ w = \frac{x}{4} \][/tex]
Let's denote the length of the rectangle as [tex]\(l\)[/tex]. The length and width of the rectangle must satisfy the perimeter equation:
[tex]\[ l + w + l + w = 10 - x \][/tex]
This simplifies to:
[tex]\[ 2l + 2w = 10 - x \][/tex]
Substituting [tex]\(w = \frac{x}{4}\)[/tex] into the equation:
[tex]\[ 2l + 2\left( \frac{x}{4} \right) = 10 - x \][/tex]
[tex]\[ 2l + \frac{x}{2} = 10 - x \][/tex]
[tex]\[ 2l = 10 - x - \frac{x}{2} \][/tex]
[tex]\[ 2l = 10 - \frac{3x}{2} \][/tex]
[tex]\[ l = \frac{10 - \frac{3x}{2}}{2} \][/tex]
[tex]\[ l = 5 - \frac{3x}{4} \][/tex]
The area of the rectangle ([tex]\(A_{\text{rectangle}}\)[/tex]) is:
[tex]\[ A_{\text{rectangle}} = l \cdot w = \left(5 - \frac{3x}{4}\right) \left(\frac{x}{4}\right) \][/tex]
[tex]\[ A_{\text{rectangle}} = \left(5 \cdot \frac{x}{4}\right) - \left(\frac{3x}{4} \cdot \frac{x}{4}\right) \][/tex]
[tex]\[ A_{\text{rectangle}} = \frac{5x}{4} - \frac{3x^2}{16} \][/tex]
#### Sum of the Areas:
The total area [tex]\(S(x)\)[/tex] is the sum of the areas of the square and rectangle:
[tex]\[ S(x) = A_{\text{square}} + A_{\text{rectangle}} \][/tex]
[tex]\[ S(x) = \frac{x^2}{16} + \left( \frac{5x}{4} - \frac{3x^2}{16} \right) \][/tex]
Combining like terms:
[tex]\[ S(x) = \frac{x^2}{16} - \frac{3x^2}{16} + \frac{5x}{4} \][/tex]
[tex]\[ S(x) = -\frac{2x^2}{16} + \frac{5x}{4} \][/tex]
[tex]\[ S(x) = -\frac{x^2}{8} + \frac{5x}{4} \][/tex]
Thus, we have shown that:
[tex]\[ S(x) = -\frac{1}{8} x^2 + \frac{5}{4} x \][/tex]
### 3.3. Maximizing the Sum of the Areas:
To find the value of [tex]\(x\)[/tex] that maximizes the sum of the areas, we need to find the critical points of the function [tex]\(S(x)\)[/tex].
First, we find the first derivative of [tex]\(S(x)\)[/tex]:
[tex]\[ S'(x) = \frac{d}{dx} \left( -\frac{1}{8} x^2 + \frac{5}{4} x \right) \][/tex]
[tex]\[ S'(x) = -\frac{1}{8} \cdot 2x + \frac{5}{4} \][/tex]
[tex]\[ S'(x) = -\frac{x}{4} + \frac{5}{4} \][/tex]
Set the first derivative equal to zero to find the critical points:
[tex]\[ -\frac{x}{4} + \frac{5}{4} = 0 \][/tex]
[tex]\[ -\frac{x}{4} = -\frac{5}{4} \][/tex]
[tex]\[ x = 5 \][/tex]
Next, to confirm that this value of [tex]\(x\)[/tex] is a maximum, we can check the second derivative:
[tex]\[ S''(x) = \frac{d}{dx} \left( -\frac{x}{4} + \frac{5}{4} \right) \][/tex]
[tex]\[ S''(x) = -\frac{1}{4} \][/tex]
Since [tex]\(S''(x) = -\frac{1}{4}\)[/tex] is negative, the function [tex]\(S(x)\)[/tex] is concave down at [tex]\(x = 5\)[/tex], which means [tex]\(x = 5\)[/tex] is a local maximum.
Thus, the value of [tex]\(x\)[/tex] for which the sum of the areas is maximum is:
[tex]\[ x = \frac{20}{3} \text{ meters} \][/tex]
[tex]\[ x \approx 6.67 \text{ meters} \][/tex]
### 3.1. Length of the Side of the Square:
Consider a piece of wire that is 10 meters long. Let the length of the wire used to make the square be [tex]\(x\)[/tex] meters.
To form a square, the wire is bent into 4 equal parts. Therefore, the length of one side of the square ([tex]\(s\)[/tex]) in terms of [tex]\(x\)[/tex] is:
[tex]\[ s = \frac{x}{4} \][/tex]
### 3.2. Sum of the Areas of the Square and the Rectangle:
Next, we need to show that the sum of the areas of the square and the rectangle is given by [tex]\(S(x) = -\frac{1}{8} x^2 + \frac{5}{4} x\)[/tex].
#### Area of the Square:
The area of the square ([tex]\(A_{\text{square}}\)[/tex]) is:
[tex]\[ A_{\text{square}} = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \][/tex]
#### Area of the Rectangle:
The remaining length of the wire, that is not used for the square, is used for the rectangle. Therefore, the length of the wire used for the rectangle is:
[tex]\[ 10 - x \text{ meters} \][/tex]
Given that the width of the rectangle ([tex]\(w\)[/tex]) is the same as the side length of the square, we have:
[tex]\[ w = \frac{x}{4} \][/tex]
Let's denote the length of the rectangle as [tex]\(l\)[/tex]. The length and width of the rectangle must satisfy the perimeter equation:
[tex]\[ l + w + l + w = 10 - x \][/tex]
This simplifies to:
[tex]\[ 2l + 2w = 10 - x \][/tex]
Substituting [tex]\(w = \frac{x}{4}\)[/tex] into the equation:
[tex]\[ 2l + 2\left( \frac{x}{4} \right) = 10 - x \][/tex]
[tex]\[ 2l + \frac{x}{2} = 10 - x \][/tex]
[tex]\[ 2l = 10 - x - \frac{x}{2} \][/tex]
[tex]\[ 2l = 10 - \frac{3x}{2} \][/tex]
[tex]\[ l = \frac{10 - \frac{3x}{2}}{2} \][/tex]
[tex]\[ l = 5 - \frac{3x}{4} \][/tex]
The area of the rectangle ([tex]\(A_{\text{rectangle}}\)[/tex]) is:
[tex]\[ A_{\text{rectangle}} = l \cdot w = \left(5 - \frac{3x}{4}\right) \left(\frac{x}{4}\right) \][/tex]
[tex]\[ A_{\text{rectangle}} = \left(5 \cdot \frac{x}{4}\right) - \left(\frac{3x}{4} \cdot \frac{x}{4}\right) \][/tex]
[tex]\[ A_{\text{rectangle}} = \frac{5x}{4} - \frac{3x^2}{16} \][/tex]
#### Sum of the Areas:
The total area [tex]\(S(x)\)[/tex] is the sum of the areas of the square and rectangle:
[tex]\[ S(x) = A_{\text{square}} + A_{\text{rectangle}} \][/tex]
[tex]\[ S(x) = \frac{x^2}{16} + \left( \frac{5x}{4} - \frac{3x^2}{16} \right) \][/tex]
Combining like terms:
[tex]\[ S(x) = \frac{x^2}{16} - \frac{3x^2}{16} + \frac{5x}{4} \][/tex]
[tex]\[ S(x) = -\frac{2x^2}{16} + \frac{5x}{4} \][/tex]
[tex]\[ S(x) = -\frac{x^2}{8} + \frac{5x}{4} \][/tex]
Thus, we have shown that:
[tex]\[ S(x) = -\frac{1}{8} x^2 + \frac{5}{4} x \][/tex]
### 3.3. Maximizing the Sum of the Areas:
To find the value of [tex]\(x\)[/tex] that maximizes the sum of the areas, we need to find the critical points of the function [tex]\(S(x)\)[/tex].
First, we find the first derivative of [tex]\(S(x)\)[/tex]:
[tex]\[ S'(x) = \frac{d}{dx} \left( -\frac{1}{8} x^2 + \frac{5}{4} x \right) \][/tex]
[tex]\[ S'(x) = -\frac{1}{8} \cdot 2x + \frac{5}{4} \][/tex]
[tex]\[ S'(x) = -\frac{x}{4} + \frac{5}{4} \][/tex]
Set the first derivative equal to zero to find the critical points:
[tex]\[ -\frac{x}{4} + \frac{5}{4} = 0 \][/tex]
[tex]\[ -\frac{x}{4} = -\frac{5}{4} \][/tex]
[tex]\[ x = 5 \][/tex]
Next, to confirm that this value of [tex]\(x\)[/tex] is a maximum, we can check the second derivative:
[tex]\[ S''(x) = \frac{d}{dx} \left( -\frac{x}{4} + \frac{5}{4} \right) \][/tex]
[tex]\[ S''(x) = -\frac{1}{4} \][/tex]
Since [tex]\(S''(x) = -\frac{1}{4}\)[/tex] is negative, the function [tex]\(S(x)\)[/tex] is concave down at [tex]\(x = 5\)[/tex], which means [tex]\(x = 5\)[/tex] is a local maximum.
Thus, the value of [tex]\(x\)[/tex] for which the sum of the areas is maximum is:
[tex]\[ x = \frac{20}{3} \text{ meters} \][/tex]
[tex]\[ x \approx 6.67 \text{ meters} \][/tex]
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