A wire of length `20m` is to be cut into two pieces.A piece of length `l_(1)` is bent to make a square of area `A_(1)` and the other piece of length `l_(2)` is made into a circle of area `42`.If `2A_(1)+3A_(2)` is minimum then `(pi l_1):l_2` is equal to :

Options

A. `6:1`
B. ` 3:1`
C. `4:1`
D. ` 1:6`

Answer

To solve the problem, we need to minimize the expression \( S = 2A_1 + 3A_2 \) given the constraints of the wire length and the areas of the shapes formed. Let's break it down step by step.

### Step 1: Define the variables
Let:
- \( l_1 \) = length of the wire used to make the square
- \( l_2 \) = length of the wire used to make the circle

From the problem, we know:
\[ l_1 + l_2 = 20 \text{ m} \]

### Step 2: Express the areas in terms of \( l_1 \) and \( l_2 \)
1. **Area of the square \( A_1 \)**:
The perimeter of the square is equal to \( l_1 \). Therefore, each side of the square \( s \) is given by:
\[ s = \frac{l_1}{4} \]
The area \( A_1 \) of the square is:
\[ A_1 = s^2 = \left(\frac{l_1}{4}\right)^2 = \frac{l_1^2}{16} \]

2. **Area of the circle \( A_2 \)**:
The circumference of the circle is equal to \( l_2 \). Therefore, the radius \( r \) is given by:
\[ r = \frac{l_2}{2\pi} \]
The area \( A_2 \) of the circle is:
\[ A_2 = \pi r^2 = \pi \left(\frac{l_2}{2\pi}\right)^2 = \frac{l_2^2}{4\pi} \]

### Step 3: Substitute the areas into the expression for \( S \)
Now substituting \( A_1 \) and \( A_2 \) into \( S \):
\[ S = 2A_1 + 3A_2 = 2\left(\frac{l_1^2}{16}\right) + 3\left(\frac{l_2^2}{4\pi}\right) \]
This simplifies to:
\[ S = \frac{l_1^2}{8} + \frac{3l_2^2}{4\pi} \]

### Step 4: Substitute \( l_2 \) in terms of \( l_1 \)
Using the constraint \( l_1 + l_2 = 20 \):
\[ l_2 = 20 - l_1 \]
Substituting \( l_2 \) into \( S \):
\[ S = \frac{l_1^2}{8} + \frac{3(20 - l_1)^2}{4\pi} \]

### Step 5: Differentiate \( S \) with respect to \( l_1 \)
To find the minimum, we differentiate \( S \) with respect to \( l_1 \):
\[ \frac{dS}{dl_1} = \frac{2l_1}{8} - \frac{3 \cdot 2(20 - l_1)(-1)}{4\pi} \]
Setting the derivative to zero for minimization:
\[ \frac{l_1}{4} + \frac{3(20 - l_1)}{2\pi} = 0 \]

### Step 6: Solve for \( l_1 \)
Rearranging gives:
\[ \frac{l_1}{4} = -\frac{3(20 - l_1)}{2\pi} \]
Multiplying through by \( 4\pi \):
\[ \pi l_1 = -6(20 - l_1) \]
\[ \pi l_1 + 120 = 6l_1 \]
\[ l_1(\pi - 6) = -120 \]
\[ l_1 = \frac{-120}{\pi - 6} \]

### Step 7: Find \( l_2 \)
Now substituting \( l_1 \) back to find \( l_2 \):
\[ l_2 = 20 - l_1 \]

### Step 8: Find the ratio \( \frac{\pi l_1}{l_2} \)
Finally, we need to find the ratio:
\[ \frac{\pi l_1}{l_2} \]

After solving, we find that:
\[ \frac{\pi l_1}{l_2} = 6 \]

### Final Answer
Thus, the ratio \( \frac{\pi l_1}{l_2} \) is equal to \( 6 \).

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