The volume of a cube is increasing at the rate of `9cm^(3)//sec`. The rate `("in "cm^(2)//sec)` at which the surface area is increasing when the edge of the cube is 9 cm, is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the rate at which the surface area of a cube is increasing given that the volume is increasing at a rate of \( 9 \, \text{cm}^3/\text{sec} \) when the edge of the cube is \( 9 \, \text{cm} \).

### Step-by-Step Solution:

1. **Understand the volume of the cube**:
The volume \( V \) of a cube with edge length \( x \) is given by:
\[ V = x^3 \]

2. **Differentiate the volume with respect to time**:
We know that the volume is changing with respect to time, so we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = 3x^2 \frac{dx}{dt} \]
Given that \( \frac{dV}{dt} = 9 \, \text{cm}^3/\text{sec} \), we can set up the equation:
\[ 9 = 3x^2 \frac{dx}{dt} \]

3. **Solve for \( \frac{dx}{dt} \)**:
Rearranging the equation to solve for \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{9}{3x^2} = \frac{3}{x^2} \]

4. **Substitute \( x = 9 \, \text{cm} \)**:
Now, we substitute \( x = 9 \, \text{cm} \) into the equation:
\[ \frac{dx}{dt} = \frac{3}{9^2} = \frac{3}{81} = \frac{1}{27} \, \text{cm/sec} \]

5. **Find the surface area of the cube**:
The surface area \( S \) of a cube is given by:
\[ S = 6x^2 \]

6. **Differentiate the surface area with respect to time**:
We differentiate \( S \) with respect to \( t \):
\[ \frac{dS}{dt} = 12x \frac{dx}{dt} \]

7. **Substitute \( x = 9 \, \text{cm} \) and \( \frac{dx}{dt} \)**:
Now substitute \( x = 9 \, \text{cm} \) and \( \frac{dx}{dt} = \frac{1}{27} \, \text{cm/sec} \):
\[ \frac{dS}{dt} = 12 \cdot 9 \cdot \frac{1}{27} \]

8. **Calculate \( \frac{dS}{dt} \)**:
\[ \frac{dS}{dt} = 12 \cdot 9 \cdot \frac{1}{27} = \frac{108}{27} = 4 \, \text{cm}^2/\text{sec} \]

### Final Answer:
The rate at which the surface area is increasing when the edge of the cube is \( 9 \, \text{cm} \) is \( 4 \, \text{cm}^2/\text{sec} \).