The surface area of a cube is increasing at the rate of `2 cm^(2)//sec`. When its edge is 90 cm, the volume is increasing at the rate of

Options

A. `1620 cm^(3)//sec`
B. `810 cm^(3)//sec`
C. `405 cm^(3)//sec`
D. `45 cm^(3)//sec`

Explanation

To solve the problem step by step, we need to find the rate of change of the volume of a cube given that the surface area is increasing at a certain rate. Here’s how we can approach it:

### Step 1: Understand the given information
- The surface area of a cube is increasing at the rate of \( \frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec} \).
- The edge of the cube \( a = 90 \, \text{cm} \).

### Step 2: Write the formulas for surface area and volume of a cube
- The surface area \( S \) of a cube is given by:
\[ S = 6a^2 \]
- The volume \( V \) of a cube is given by:
\[ V = a^3 \]

### Step 3: Differentiate the surface area with respect to time
- Differentiate the surface area formula with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(6a^2) = 12a \frac{da}{dt} \]

### Step 4: Substitute the known rate of change of surface area
- We know \( \frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec} \):
\[ 2 = 12a \frac{da}{dt} \]

### Step 5: Solve for \( \frac{da}{dt} \)
- Substitute \( a = 90 \, \text{cm} \) into the equation:
\[ 2 = 12 \cdot 90 \cdot \frac{da}{dt} \]
\[ 2 = 1080 \frac{da}{dt} \]
\[ \frac{da}{dt} = \frac{2}{1080} = \frac{1}{540} \, \text{cm/sec} \]

### Step 6: Differentiate the volume with respect to time
- Differentiate the volume formula with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(a^3) = 3a^2 \frac{da}{dt} \]

### Step 7: Substitute \( a \) and \( \frac{da}{dt} \) into the volume differentiation
- Substitute \( a = 90 \, \text{cm} \) and \( \frac{da}{dt} = \frac{1}{540} \, \text{cm/sec} \):
\[ \frac{dV}{dt} = 3 \cdot (90)^2 \cdot \frac{1}{540} \]
\[ = 3 \cdot 8100 \cdot \frac{1}{540} \]
\[ = \frac{24300}{540} \]
\[ = 45 \, \text{cm}^3/\text{sec} \]

### Final Answer
The volume is increasing at the rate of \( 45 \, \text{cm}^3/\text{sec} \).

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