Each side of a square is increasing at the uniform rate of 1m/sec. If after sometime the area of the square is increasing at the rate of `8m^(2)`/sec, then the area of square at that time in sq. meters is

Options

A. 4
B. 9
C. 16
D. 25

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the side length and area of the square.
Let the side length of the square be denoted as \( a \). The area \( A \) of the square can be expressed as:
\[ A = a^2 \]

### Step 2: Differentiate the area with respect to time.
To find how the area changes with time, we differentiate \( A \) with respect to \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(a^2) = 2a \frac{da}{dt} \]
where \( \frac{da}{dt} \) is the rate of change of the side length.

### Step 3: Substitute the known values.
We know that \( \frac{da}{dt} = 1 \, \text{m/s} \) (the side length is increasing at a rate of 1 m/s) and that \( \frac{dA}{dt} = 8 \, \text{m}^2/\text{s} \) (the area is increasing at a rate of 8 m²/s). Substituting these values into the differentiated equation gives:
\[ 8 = 2a \cdot 1 \]

### Step 4: Solve for the side length \( a \).
From the equation \( 8 = 2a \), we can solve for \( a \):
\[ 2a = 8 \implies a = \frac{8}{2} = 4 \, \text{m} \]

### Step 5: Calculate the area of the square at this time.
Now that we have the side length \( a = 4 \, \text{m} \), we can find the area \( A \):
\[ A = a^2 = 4^2 = 16 \, \text{m}^2 \]

### Final Answer:
The area of the square at that time is \( 16 \, \text{m}^2 \).

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