At an instant the diagonal of a square is increasing at the rate of 0.2 cm/sec and the area is increasing at the rate of `6 cm^(2)//sec`. At the moment its side, is

Options

A. `(30)/(sqrt(2))cm`
B. `30sqrt(2)` cm
C. 30 cm
D. 15 cm

Answer

To solve the problem step by step, we will analyze the given information and use calculus to find the side of the square at the moment when the diagonal and area are changing at specified rates.

### Step-by-Step Solution:

1. **Identify Given Information:**
- The rate at which the diagonal of the square is increasing: \( \frac{dx}{dt} = 0.2 \, \text{cm/sec} \)
- The rate at which the area of the square is increasing: \( \frac{dA}{dt} = 6 \, \text{cm}^2/\text{sec} \)

2. **Relate the Diagonal and Side of the Square:**
- Let the side of the square be \( a \).
- The diagonal \( x \) of the square can be expressed in terms of the side using the Pythagorean theorem:
\[ x = a\sqrt{2} \]

3. **Differentiate the Diagonal with Respect to Time:**
- Differentiating both sides with respect to time \( t \):
\[ \frac{dx}{dt} = \sqrt{2} \frac{da}{dt} \]

4. **Relate the Area to the Side of the Square:**
- The area \( A \) of the square is given by:
\[ A = a^2 \]
- Differentiating the area with respect to time:
\[ \frac{dA}{dt} = 2a \frac{da}{dt} \]

5. **Substitute Known Values:**
- From the first differentiation, we can express \( \frac{da}{dt} \) in terms of \( \frac{dx}{dt} \):
\[ \frac{da}{dt} = \frac{1}{\sqrt{2}} \frac{dx}{dt} = \frac{1}{\sqrt{2}} \cdot 0.2 = \frac{0.2}{\sqrt{2}} \, \text{cm/sec} \]

6. **Substitute \( \frac{da}{dt} \) into the Area Equation:**
- Now substitute \( \frac{da}{dt} \) into the area equation:
\[ 6 = 2a \left( \frac{0.2}{\sqrt{2}} \right) \]
- Simplifying this gives:
\[ 6 = \frac{0.4a}{\sqrt{2}} \]

7. **Solve for \( a \):**
- Rearranging the equation to solve for \( a \):
\[ a = \frac{6\sqrt{2}}{0.4} = \frac{6\sqrt{2}}{\frac{2}{5}} = 15\sqrt{2} \, \text{cm} \]

8. **Final Result:**
- Therefore, the side of the square at that moment is:
\[ a = 15\sqrt{2} \, \text{cm} \]