https://file.aiquickdraw.com/mathbot/file/1727930323860l7mf7on2.jpg

The problem you're asking involves finding the rate of change of the area of a rectangle, given the rates of change of its width and length.

Let's break it down:

1. The width \( W \) is increasing at a rate of \( \frac{1}{2} \, \text{in/sec} \).
2. The length \( L \) is increasing at a rate of \( \frac{1}{3} \, \text{in/sec} \).
3. We are asked to find the rate at which the area \( A \) is changing when \( W = 4 \, \text{inches} \) and \( L = 10 \, \text{inches} \).

### Formula for Area of a Rectangle:
The area \( A \) of a rectangle is given by:
\[
A = L \times W
\]
We need to find \( \frac{dA}{dt} \), the rate of change of the area with respect to time.

### Using the Product Rule:
To find \( \frac{dA}{dt} \), we differentiate \( A = L \times W \) with respect to time \( t \):
\[
\frac{dA}{dt} = L \frac{dW}{dt} + W \frac{dL}{dt}
\]
Where:
- \( \frac{dW}{dt} = \frac{1}{2} \, \text{in/sec} \) (rate of change of width)
- \( \frac{dL}{dt} = \frac{1}{3} \, \text{in/sec} \) (rate of change of length)

### Substituting the Values:
Given \( L = 10 \, \text{inches} \) and \( W = 4 \, \text{inches} \), we substitute into the equation:
\[
\frac{dA}{dt} = (10)(\frac{1}{2}) + (4)(\frac{1}{3})
\]
Simplify the terms:
\[
\frac{dA}{dt} = 5 + \frac{4}{3}
\]
Convert \( 5 \) to a fraction:
\[
\frac{dA}{dt} = \frac{15}{3} + \frac{4}{3} = \frac{19}{3} \, \text{in}^2/\text{sec}
\]

### Conclusion:
The rate of change of the area is \( \frac{19}{3} \, \text{in}^2/\text{sec} \), so the correct answer is the option:
\[
\boxed{\frac{19}{3} \, \text{in}^2/\text{sec}}
\]

Would you like any more details or clarifications? 

Here are some related questions:
1. How would the answer change if the rates of change of width and length were reversed?
2. What happens if both the width and length are decreasing instead of increasing?
3. How would you apply this method to find the rate of change of the perimeter?
4. What is the geometric interpretation of the product rule in this context?
5. How would the formula change if the rectangle had variable dimensions over time?

**Tip**: When dealing with rates of change in geometry, always remember to apply the chain rule or product rule carefully to account for all the variables changing with time.

The sides of a rectangle are changing in a way that the width is increasing at the rate of 1/2 inches per second, and the length is increasing at a rate of 1/3 inches per second. At what rate is the area of this rectangle changing when the width is 4 inches and the length is 10 inches?

Learn how to solve a related rates problem involving the rate of change of the area of a rectangle. This problem involves the product rule and differentiation, where the width is increasing at 1/2 inches per second and the length at 1/3 inches per second.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation