The volume of a cube is increasing at the rate of `18 cm^(3)` per second. When the edge of the cube is 12 cm, then the rate in `cm^(2)//s`, at which the surface area of the cube increases, is

Answer

To solve the problem step by step, we will follow the given information and apply the necessary calculus concepts.

### Step 1: Understand the relationship between volume and edge length of the cube.
The volume \( V \) of a cube with edge length \( r \) is given by the formula:
\[ V = r^3 \]

### Step 2: Differentiate the volume with respect to time \( t \).
To find how the volume changes with respect to time, we differentiate both sides of the volume equation:
\[ \frac{dV}{dt} = 3r^2 \frac{dr}{dt} \]

### Step 3: Substitute the known values.
We know from the problem that:
- \( \frac{dV}{dt} = 18 \, \text{cm}^3/\text{s} \)
- \( r = 12 \, \text{cm} \)

Substituting these values into the differentiated equation gives:
\[ 18 = 3(12^2) \frac{dr}{dt} \]
Calculating \( 12^2 \):
\[ 12^2 = 144 \]
Thus, we have:
\[ 18 = 3 \times 144 \frac{dr}{dt} \]
\[ 18 = 432 \frac{dr}{dt} \]

### Step 4: Solve for \( \frac{dr}{dt} \).
To find \( \frac{dr}{dt} \), we rearrange the equation:
\[ \frac{dr}{dt} = \frac{18}{432} = \frac{1}{24} \, \text{cm/s} \]

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

### Step 6: Differentiate the surface area with respect to time \( t \).
Differentiating the surface area with respect to time gives:
\[ \frac{dS}{dt} = 12r \frac{dr}{dt} \]

### Step 7: Substitute the known values into the differentiated surface area equation.
Using \( r = 12 \, \text{cm} \) and \( \frac{dr}{dt} = \frac{1}{24} \, \text{cm/s} \):
\[ \frac{dS}{dt} = 12(12) \left(\frac{1}{24}\right) \]
Calculating \( 12 \times 12 \):
\[ 12 \times 12 = 144 \]
Thus, we have:
\[ \frac{dS}{dt} = 144 \left(\frac{1}{24}\right) = 6 \, \text{cm}^2/\text{s} \]

### Conclusion:
The rate at which the surface area of the cube is increasing is:
\[ \frac{dS}{dt} = 6 \, \text{cm}^2/\text{s} \]