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 when the edge length is 9 cm, given that the volume of the cube is increasing at a rate of 9 cm³/sec.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( a \) be the length of the edge of the cube.
- The volume \( V \) of the cube is given by the formula:
\[ V = a^3 \]
- The surface area \( S \) of the cube is given by the formula:
\[ S = 6a^2 \]

2. **Differentiate Volume with Respect to Time**:
- To find the rate of change of volume with respect to time, we differentiate \( V \):
\[ \frac{dV}{dt} = 3a^2 \frac{da}{dt} \]
- We know from the problem that \( \frac{dV}{dt} = 9 \, \text{cm}^3/\text{sec} \).

3. **Substitute Known Values**:
- Substitute \( \frac{dV}{dt} \) into the differentiated volume equation:
\[ 9 = 3a^2 \frac{da}{dt} \]
- When \( a = 9 \, \text{cm} \):
\[ 9 = 3(9^2) \frac{da}{dt} \]
- Simplifying gives:
\[ 9 = 3 \times 81 \frac{da}{dt} \implies 9 = 243 \frac{da}{dt} \]
- Solving for \( \frac{da}{dt} \):
\[ \frac{da}{dt} = \frac{9}{243} = \frac{1}{27} \, \text{cm/sec} \]

4. **Differentiate Surface Area with Respect to Time**:
- Now, we differentiate the surface area \( S \):
\[ \frac{dS}{dt} = 12a \frac{da}{dt} \]

5. **Substitute Known Values into Surface Area Rate**:
- Substitute \( a = 9 \, \text{cm} \) and \( \frac{da}{dt} = \frac{1}{27} \, \text{cm/sec} \):
\[ \frac{dS}{dt} = 12(9) \left(\frac{1}{27}\right) \]
- Simplifying gives:
\[ \frac{dS}{dt} = 108 \cdot \frac{1}{27} = 4 \, \text{cm}^2/\text{sec} \]

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