A cube has an edge of 2 feet. The edge is increasing at the rate of 4 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed. Hint: Remember that the volume of a cube is the cube (third power) of the length of a side. V (m) = 自国 feer
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![### Volume of a Cube with Variable Edge Length
**Topic: Mathematics**
**Subtopic: Calculus**
**Module: Calculating Derivatives and Integrals**
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#### Problem:
A cube has an edge of 2 feet. The edge is increasing at the rate of 4 feet per minute. Express the volume of the cube as a function of \( m \), the number of minutes elapsed.
#### Hint:
Remember that the volume of a cube is the cube (third power) of the length of a side.
#### Solution:
Let \( s \) be the length of the edge of the cube in feet after \( m \) minutes. Initially, \( s = 2 \) feet.
Given that the edge is increasing at the rate of 4 feet per minute, the length of the edge after \( m \) minutes is:
\[ s = 2 + 4m \]
The volume \( V \) of a cube with edge length \( s \) is given by:
\[ V = s^3 \]
Therefore, the volume as a function of \( m \) is:
\[ V(m) = (2 + 4m)^3 \]
\[ V(m) = 8 (1 + 2m)^3 \, \text{cubic feet} \]
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This content is designed to help students understand how to set up and solve problems involving the rates of change and the calculation of volumes in calculus. For further assistance, please refer to the Academic Support resources or consult your course instructor.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21bd8fd9-92a3-41b3-9bf9-2b1154ed45ae%2F60df9f3d-78f7-4b9f-b450-dfca6ade1071%2Fy7qnrz7_processed.png&w=3840&q=75)
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