A rectangle has a length that is decreasing at a rate of 11 mm per second with the width being held constant. What is the rate of change of the area of the rectangle if the width is 9 mm? 

To determine the rate of change of the area of the rectangle, we can use the formula for the area of a rectangle:

\[
A = \text{length} \times \text{width}
\]

Let \( l \) represent the length and \( w \) represent the width of the rectangle. Then, the area \( A \) can be expressed as:

\[
A = l \times w
\]

Given:
- The width \( w = 9 \) mm (constant)
- The rate of change of the length \( \frac{dl}{dt} = -11 \) mm per second (negative because the length is decreasing)

We need to find the rate of change of the area, \( \frac{dA}{dt} \).

Differentiate the area formula with respect to time \( t \):

\[
\frac{dA}{dt} = \frac{dl}{dt} \times w + l \times \frac{dw}{dt}
\]

Since the width \( w \) is constant, \( \frac{dw}{dt} = 0 \). Therefore, the equation simplifies to:

\[
\frac{dA}{dt} = \frac{dl}{dt} \times w
\]

Substitute the given values:

\[
\frac{dA}{dt} = (-11) \times 9 = -99 \text{ mm}^2 \text{ per second}
\]

Thus, the rate of change of the area of the rectangle is \(-99\) square millimeters per second.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How would the rate of change of the area be different if the width was also changing?
2. What if the length was increasing instead of decreasing? How would that affect the rate?
3. How does the rate of change of the area compare if the width was larger or smaller?
4. What would happen to the area if both the length and width were decreasing?
5. How would you calculate the rate of change of the perimeter of the rectangle?

**Tip:** When calculating the rate of change of an area, it's important to remember that only the variables that are changing contribute to the derivative.

Explore how to calculate the rate of change of the area of a rectangle when the length decreases at 11 mm per second and the width remains constant at 9 mm. Understand the derivative formula used and its application in solving related problems.

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