To find the rate of change of the area of a rectangle whose length and width are changing over time, we can use related rates in calculus.

Given:

• Length of the rectangle, $x$, is decreasing at a rate of $\frac{dx}{dt} = -5$ cm/min.
• Width of the rectangle, $y$, is increasing at a rate of $\frac{dy}{dt} = 4$ cm/min.
• At the moment we are considering, $x = 8$ cm and $y = 6$ cm.

Objective:

Find the rate of change of the area, $\frac{dA}{dt}$, when $x = 8$ cm and $y = 6$ cm.

Solution:

1. Write the formula for the area of the rectangle: $A = x \cdot y$

2. Differentiate both sides with respect to time $t$: Using the product rule, we get: $\frac{dA}{dt} = \frac{dx}{dt} \cdot y + x \cdot \frac{dy}{dt}$

3. Substitute the known values:

   • $\frac{dx}{dt} = -5$ cm/min,
   • $\frac{dy}{dt} = 4$ cm/min,
   • $x = 8$ cm,
   • $y = 6$ cm.

   Substitute these values into the equation: $\frac{dA}{dt} = (-5)(6) + (8)(4)$ $\frac{dA}{dt} = -30 + 32$ $\frac{dA}{dt} = 2$

Final Answer:

The rate of change of the area of the rectangle is $2$ cm²/min.

So, $\frac{dA}{dt} = 2$ cm²/min.

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Would you like more clarification on related rates or additional practice problems? Here are some related questions:

1. How do related rates apply to other geometric shapes?
2. Can you find the rate of change of the perimeter in this scenario?
3. What if both the length and width were decreasing? How would that affect $\frac{dA}{dt}$?
4. How does the product rule help in related rates problems?
5. What is the importance of signs (positive/negative) in interpreting rates?

Tip: Always pay attention to the direction of change when solving related rates problems.