If the edge of a cube increases at the rate of 60 cm per second, at what rate the volume is increasing when the edge is 90 cm

Options

A. 486000 cu cm per sec
B. 1458000cu cm per sec
C. 43740000cu cm per sec
D. none of these

Explanation

To solve the problem, we need to find the rate at which the volume of a cube is increasing when the edge of the cube is 90 cm, given that the edge is increasing at a rate of 60 cm/s.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( A \) be the length of the edge of the cube.
- The volume \( V \) of the cube is given by the formula:
\[ V = A^3 \]
- We know that \( \frac{dA}{dt} = 60 \, \text{cm/s} \).

2. **Differentiate the Volume with Respect to Time**:
- To find the rate of change of volume with respect to time, we differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(A^3) \]
- Using the chain rule, we get:
\[ \frac{dV}{dt} = 3A^2 \frac{dA}{dt} \]

3. **Substitute the Known Values**:
- We need to find \( \frac{dV}{dt} \) when \( A = 90 \, \text{cm} \) and \( \frac{dA}{dt} = 60 \, \text{cm/s} \).
- Substitute these values into the differentiated equation:
\[ \frac{dV}{dt} = 3(90^2)(60) \]

4. **Calculate \( 90^2 \)**:
- First, calculate \( 90^2 \):
\[ 90^2 = 8100 \]

5. **Calculate \( \frac{dV}{dt} \)**:
- Now substitute \( 90^2 \) back into the equation:
\[ \frac{dV}{dt} = 3 \times 8100 \times 60 \]
- Calculate \( 3 \times 8100 \):
\[ 3 \times 8100 = 24300 \]
- Now multiply by 60:
\[ 24300 \times 60 = 1458000 \]

6. **Final Answer**:
- Therefore, the rate at which the volume is increasing when the edge is 90 cm is:
\[ \frac{dV}{dt} = 1458000 \, \text{cm}^3/\text{s} \]

### Summary:
The volume of the cube is increasing at a rate of **1458000 cm³/s** when the edge is 90 cm.