Question
How do I maximize and minimize the combined area of a square and equilateral triangle formed by cutting a 12-meter wire into two parts?
Answers
GPT-4o mini
To maximize and minimize the combined area of a square and an equilateral triangle formed by cutting a 12-meter wire into two parts, let us denote:
- \( x \) as the length of the wire used for the square.
- \( 12 - x \) as the length of the wire used for the equilateral triangle.
### Step 1: Formulate the area
1. **Area of the square:**
The perimeter of the square is \( x \). Therefore, each side length \( s \) of the square is:
\[
s = \frac{x}{4}
\]
The area \( A_s \) of the square is:
\[
A_s = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}
\]
2. **Area of the equilateral triangle:**
The perimeter of the equilateral triangle is \( 12 - x \). Each side length \( t \) of the triangle is:
\[
t = \frac{12 - x}{3}
\]
The area \( A_t \) of the equilateral triangle can be computed using the formula:
\[
A_t = \frac{\sqrt{3}}{4} t^2 = \frac{\sqrt{3}}{4} \left(\frac{12 - x}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{(12 - x)^2}{9} = \frac{\sqrt{3}(12 - x)^2}{36}
\]
### Step 2: Combined area
The total area \( A \) as a function of \( x \) is given by the sum of the areas of the square and triangle:
\[
A(x) = A_s + A_t = \frac{x^2}{16} + \frac{\sqrt{3}(12 - x)^2}{36}
\]
### Step 3: Find Extremes
To find the maximum and minimum areas, we differentiate \( A(x) \) with respect to \( x \) and set the derivative equal to zero. First, we simplify the expression for \( A(x) \):
\[
A(x) = \frac{x^2}{16} + \frac{\sqrt{3}(144 - 24x + x^2)}{36}
\]
To make differentiation easier, let's express both terms with a common denominator:
The common denominator of 16 and 36 is 144. Thus, we can rewrite the area as:
\[
A(x) = \frac{9x^2}{144} + \frac{4\sqrt{3}(144 - 24x + x^2)}{144}
\]
Combining the terms, we have:
\[
A(x) = \frac{9x^2 + 576 - 96\sqrt{3}x + 4\sqrt{3}x^2}{144} = \frac{(9 + 4\sqrt{3})x^2 - 96\sqrt{3}x + 576}{144}
\]
### Step 4: Differentiate \( A(x) \)
Now, differentiate \( A(x) \) with respect to \( x \):
\[
A'(x) = \frac{1}{144} \left((9 + 4\sqrt{3})\cdot 2x - 96\sqrt{3}\right)
\]
Setting \( A'(x) = 0 \):
\[
(9 + 4\sqrt{3})\cdot 2x - 96\sqrt{3} = 0 \Rightarrow 2x(9 + 4\sqrt{3}) = 96\sqrt{3}
\]
\[
x = \frac{96\sqrt{3}}{2(9 + 4\sqrt{3})}
\]
\[
x = \frac{48\sqrt{3}}{9 + 4\sqrt{3}} \approx 6.93
\]
### Step 5: Finding the areas
Substituting the value of \( x \) back into the area function allows us to determine the max area. We would then also evaluate areas at the endpoints of the constraint \( x = 0 \) and \( x = 12 \) to check if they give a minimum or maximum.
1. **When \( x = 0 \)**: Only the triangle is formed.
2. **When \( x = 12 \)**: Only the square is formed.
Calculate areas at these points with similar formulas.
### Step 6: Conclusion
Thus, by evaluating \( A(x) \) at the critical point \( x \approx 6.93 \) and the endpoints (0 and 12), we can identify the maximum and minimum combined area formed by the square and triangle.
- \( x \) as the length of the wire used for the square.
- \( 12 - x \) as the length of the wire used for the equilateral triangle.
### Step 1: Formulate the area
1. **Area of the square:**
The perimeter of the square is \( x \). Therefore, each side length \( s \) of the square is:
\[
s = \frac{x}{4}
\]
The area \( A_s \) of the square is:
\[
A_s = s^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}
\]
2. **Area of the equilateral triangle:**
The perimeter of the equilateral triangle is \( 12 - x \). Each side length \( t \) of the triangle is:
\[
t = \frac{12 - x}{3}
\]
The area \( A_t \) of the equilateral triangle can be computed using the formula:
\[
A_t = \frac{\sqrt{3}}{4} t^2 = \frac{\sqrt{3}}{4} \left(\frac{12 - x}{3}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{(12 - x)^2}{9} = \frac{\sqrt{3}(12 - x)^2}{36}
\]
### Step 2: Combined area
The total area \( A \) as a function of \( x \) is given by the sum of the areas of the square and triangle:
\[
A(x) = A_s + A_t = \frac{x^2}{16} + \frac{\sqrt{3}(12 - x)^2}{36}
\]
### Step 3: Find Extremes
To find the maximum and minimum areas, we differentiate \( A(x) \) with respect to \( x \) and set the derivative equal to zero. First, we simplify the expression for \( A(x) \):
\[
A(x) = \frac{x^2}{16} + \frac{\sqrt{3}(144 - 24x + x^2)}{36}
\]
To make differentiation easier, let's express both terms with a common denominator:
The common denominator of 16 and 36 is 144. Thus, we can rewrite the area as:
\[
A(x) = \frac{9x^2}{144} + \frac{4\sqrt{3}(144 - 24x + x^2)}{144}
\]
Combining the terms, we have:
\[
A(x) = \frac{9x^2 + 576 - 96\sqrt{3}x + 4\sqrt{3}x^2}{144} = \frac{(9 + 4\sqrt{3})x^2 - 96\sqrt{3}x + 576}{144}
\]
### Step 4: Differentiate \( A(x) \)
Now, differentiate \( A(x) \) with respect to \( x \):
\[
A'(x) = \frac{1}{144} \left((9 + 4\sqrt{3})\cdot 2x - 96\sqrt{3}\right)
\]
Setting \( A'(x) = 0 \):
\[
(9 + 4\sqrt{3})\cdot 2x - 96\sqrt{3} = 0 \Rightarrow 2x(9 + 4\sqrt{3}) = 96\sqrt{3}
\]
\[
x = \frac{96\sqrt{3}}{2(9 + 4\sqrt{3})}
\]
\[
x = \frac{48\sqrt{3}}{9 + 4\sqrt{3}} \approx 6.93
\]
### Step 5: Finding the areas
Substituting the value of \( x \) back into the area function allows us to determine the max area. We would then also evaluate areas at the endpoints of the constraint \( x = 0 \) and \( x = 12 \) to check if they give a minimum or maximum.
1. **When \( x = 0 \)**: Only the triangle is formed.
2. **When \( x = 12 \)**: Only the square is formed.
Calculate areas at these points with similar formulas.
### Step 6: Conclusion
Thus, by evaluating \( A(x) \) at the critical point \( x \approx 6.93 \) and the endpoints (0 and 12), we can identify the maximum and minimum combined area formed by the square and triangle.