`l_1 and l_2` are the side lengths of two variable squares ` S_1, and S_2`, respectively. If `l_1=l_2+l_2^3+6` then rate of change of the area of `S_2`, with respect to rate of change of the area of `S_1` when `l_2 = 1` is equal to

Options

A. `3/4`
B. `4/3`
C. `3/2`
D. `1/32`

Answer

To solve the problem, we need to find the rate of change of the area of square \( S_2 \) with respect to the area of square \( S_1 \) when \( l_2 = 1 \).

### Step-by-step Solution:

1. **Define the areas of the squares**:
- The area of square \( S_1 \) is given by:
\[ A_1 = l_1^2 \]
- The area of square \( S_2 \) is given by:
\[ A_2 = l_2^2 \]

2. **Express \( l_1 \) in terms of \( l_2 \)**:
- From the problem statement, we have:
\[ l_1 = l_2 + l_2^3 + 6 \]

3. **Differentiate the areas with respect to time \( t \)**:
- Differentiate \( A_1 \):
\[ \frac{dA_1}{dt} = 2l_1 \frac{dl_1}{dt} \]
- Differentiate \( A_2 \):
\[ \frac{dA_2}{dt} = 2l_2 \frac{dl_2}{dt} \]

4. **Differentiate \( l_1 \)**:
- Differentiate \( l_1 \) with respect to \( t \):
\[ \frac{dl_1}{dt} = \frac{d}{dt}(l_2 + l_2^3 + 6) = \frac{dl_2}{dt} + 3l_2^2 \frac{dl_2}{dt} \]
- This can be simplified to:
\[ \frac{dl_1}{dt} = \left(1 + 3l_2^2\right) \frac{dl_2}{dt} \]

5. **Substitute \( l_1 \) and \( l_2 \) into the derivatives**:
- Substitute \( l_2 = 1 \):
\[ l_1 = 1 + 1^3 + 6 = 8 \]
- Now substitute \( l_2 = 1 \) into the derivatives:
\[ \frac{dA_1}{dt} = 2 \cdot 8 \cdot \left(1 + 3 \cdot 1^2\right) \frac{dl_2}{dt} = 2 \cdot 8 \cdot 4 \frac{dl_2}{dt} = 64 \frac{dl_2}{dt} \]
\[ \frac{dA_2}{dt} = 2 \cdot 1 \cdot \frac{dl_2}{dt} = 2 \frac{dl_2}{dt} \]

6. **Find the ratio of the rates of change of areas**:
- The ratio \( \frac{dA_2}{dA_1} \) is given by:
\[ \frac{dA_2}{dA_1} = \frac{2 \frac{dl_2}{dt}}{64 \frac{dl_2}{dt}} = \frac{2}{64} = \frac{1}{32} \]

### Final Answer:
The rate of change of the area of \( S_2 \) with respect to the area of \( S_1 \) when \( l_2 = 1 \) is:
\[ \frac{1}{32} \]