Areaof a square plate is increasing at the uniform rate of 2 `cm^(2)`/sec. Find the rate at which the perimeter is increasing when the side of the square is 16 cm long.

Options

A. 4 cm /sec
B. `(1)/(4)` cm /sec
C. 2 cm /sec
D. `(1)/(2)` cm /sec

Answer

To solve the problem, we need to find the rate at which the perimeter of a square is increasing given that the area is increasing at a uniform rate of 2 cm²/sec when the side of the square is 16 cm long.

### Step-by-Step Solution:

1. **Understand the relationship between area and side length**:
The area \( A \) of a square with side length \( s \) is given by:
\[ A = s^2 \]

2. **Differentiate the area with respect to time**:
To find how the area changes with time, we differentiate both sides with respect to \( t \):
\[ \frac{dA}{dt} = 2s \frac{ds}{dt} \]
Here, \( \frac{ds}{dt} \) is the rate of change of the side length with respect to time.

3. **Substitute the known rate of area change**:
We know that \( \frac{dA}{dt} = 2 \) cm²/sec (the area is increasing at this rate). Therefore, we can set up the equation:
\[ 2 = 2s \frac{ds}{dt} \]

4. **Solve for \( \frac{ds}{dt} \)**:
Rearranging the equation gives:
\[ \frac{ds}{dt} = \frac{2}{2s} = \frac{1}{s} \]

5. **Substitute \( s = 16 \) cm**:
Now, we substitute \( s = 16 \) cm into the equation:
\[ \frac{ds}{dt} = \frac{1}{16} \text{ cm/sec} \]

6. **Find the perimeter**:
The perimeter \( P \) of a square is given by:
\[ P = 4s \]

7. **Differentiate the perimeter with respect to time**:
Differentiating the perimeter with respect to time gives:
\[ \frac{dP}{dt} = 4 \frac{ds}{dt} \]

8. **Substitute \( \frac{ds}{dt} \)**:
Now substitute \( \frac{ds}{dt} = \frac{1}{16} \) cm/sec into the equation:
\[ \frac{dP}{dt} = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \text{ cm/sec} \]

### Final Answer:
The rate at which the perimeter is increasing when the side of the square is 16 cm long is:
\[ \frac{1}{4} \text{ cm/sec} \]