Math Problem Statement

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s. When the length is 10 cm and the width is 6 cm, how fast is the area of the rectangle increasing?

Solution

Given the problem's context and your question about the rate at which the area of the rectangle is increasing, let's analyze it using related rates.

Given information:

  • The length ll of the rectangle is increasing at a rate of dldt=8cm/s\frac{dl}{dt} = 8 \, \text{cm/s}.
  • The width ww of the rectangle is increasing at a rate of dwdt=3cm/s\frac{dw}{dt} = 3 \, \text{cm/s}.
  • When the length is l=10cml = 10 \, \text{cm} and the width is w=6cmw = 6 \, \text{cm}, we need to find how fast the area AA of the rectangle is increasing.

Step 1: Differentiate the area formula

The area of a rectangle is given by: A=lwA = l \cdot w

To find how fast the area is increasing, we need to differentiate AA with respect to time tt: dAdt=Aldldt+Awdwdt\frac{dA}{dt} = \frac{\partial A}{\partial l} \cdot \frac{dl}{dt} + \frac{\partial A}{\partial w} \cdot \frac{dw}{dt}

Since Al=w\frac{\partial A}{\partial l} = w and Aw=l\frac{\partial A}{\partial w} = l, the equation becomes: dAdt=wdldt+ldwdt\frac{dA}{dt} = w \cdot \frac{dl}{dt} + l \cdot \frac{dw}{dt}

Step 2: Substitute the values

Substituting the given values:

  • l=10cml = 10 \, \text{cm}
  • w=6cmw = 6 \, \text{cm}
  • dldt=8cm/s\frac{dl}{dt} = 8 \, \text{cm/s}
  • dwdt=3cm/s\frac{dw}{dt} = 3 \, \text{cm/s}

dAdt=68+103\frac{dA}{dt} = 6 \cdot 8 + 10 \cdot 3 dAdt=48+30\frac{dA}{dt} = 48 + 30 dAdt=78cm2/s\frac{dA}{dt} = 78 \, \text{cm}^2/\text{s}

Conclusion

The area of the rectangle is increasing at a rate of 78cm2/s78 \, \text{cm}^2/\text{s}.

Would you like more details or have any questions about this calculation?

Here are five related questions to further explore this topic:

  1. What happens to the rate of area change if the width's rate of increase doubles?
  2. How does the formula change if only the length of the rectangle increases while the width stays constant?
  3. What is the geometric interpretation of related rates in this context?
  4. How can we apply this method to other shapes, like triangles or circles?
  5. What if the length and width were decreasing instead of increasing—how would that affect the rate of area change?

Tip: In related rates problems, it's helpful to differentiate all quantities with respect to time before substituting values to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Related Rates
Differentiation
Geometry

Formulas

Area of a rectangle: A = l * w
Derivative of area with respect to time: dA/dt = w * (dl/dt) + l * (dw/dt)

Theorems

Chain Rule of Differentiation

Suitable Grade Level

Grades 10-12