A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Options

A. `L_1= 112/(pi+4)`, `L_1= (28pi)/(pi+4)`
B. `L_1= 142/(pi+4)`, `L_1= (28pi)/(pi+4)`
C. `L_1= 112/(pi+4)`, `L_1= pi/(pi+4)`
D. None of these

Explanation

To solve the problem of cutting a wire of length 28 m into two pieces to minimize the combined area of a square and a circle, we can follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = length of the wire used for the square
- \( y \) = length of the wire used for the circle

From the problem, we know:
\[ x + y = 28 \]

### Step 2: Express Areas
1. **Area of the Square**:
The perimeter of the square is \( x \), so the side length \( a \) of the square is:
\[ a = \frac{x}{4} \]
Therefore, the area \( A_s \) of the square is:
\[ A_s = a^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \]

2. **Area of the Circle**:
The circumference of the circle is \( y \), so the radius \( r \) of the circle is:
\[ r = \frac{y}{2\pi} \]
Therefore, the area \( A_c \) of the circle is:
\[ A_c = \pi r^2 = \pi \left(\frac{y}{2\pi}\right)^2 = \frac{y^2}{4\pi} \]

### Step 3: Total Area
The total area \( A \) is the sum of the areas of the square and the circle:
\[ A = A_s + A_c = \frac{x^2}{16} + \frac{y^2}{4\pi} \]

### Step 4: Substitute for One Variable
Using the constraint \( y = 28 - x \), we can substitute \( y \) into the area equation:
\[ A = \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi} \]

### Step 5: Simplify the Area Function
Now, expand and simplify:
\[ A = \frac{x^2}{16} + \frac{(784 - 56x + x^2)}{4\pi} \]
\[ A = \frac{x^2}{16} + \frac{784}{4\pi} - \frac{56x}{4\pi} + \frac{x^2}{4\pi} \]
Combine the \( x^2 \) terms:
\[ A = \left(\frac{1}{16} + \frac{1}{4\pi}\right)x^2 - \frac{14x}{\pi} + \frac{196}{\pi} \]

### Step 6: Differentiate the Area Function
To find the minimum area, we differentiate \( A \) with respect to \( x \):
\[ \frac{dA}{dx} = 2\left(\frac{1}{16} + \frac{1}{4\pi}\right)x - \frac{14}{\pi} \]

### Step 7: Set Derivative to Zero
Set the derivative equal to zero to find critical points:
\[ 2\left(\frac{1}{16} + \frac{1}{4\pi}\right)x - \frac{14}{\pi} = 0 \]
Solving for \( x \):
\[ x = \frac{14/\pi}{2\left(\frac{1}{16} + \frac{1}{4\pi}\right)} \]

### Step 8: Calculate \( y \)
Once \( x \) is found, calculate \( y \):
\[ y = 28 - x \]

### Step 9: Conclusion
The lengths of the two pieces of wire that minimize the combined area of the square and the circle can be determined from the values of \( x \) and \( y \).