Question

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 24 m per year at a time when the region is 300 m wide. How fast is the area changing at that point in time

Answer 1

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

Geometry

• Area of a Rectangle: A = lw

Algebra I

• Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

1. Substitute in l:                    A = (2w)w
2. Multiply:                              A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

1. Differentiate [Basic Power Rule]:                                                                   $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
2. Simplify:                                                                                                           $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

1. Substitute:                    $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
2. Multiply:                        $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
3. Multiply:                        $\frac{dA}{dt} = 28800 \ m^2/year$

Answer 2

Answer:

28,800 m²/yr

Explanation:

This rectangle has dimensions such that:

• width = w
• length = 2w  

We are given $\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}$ and want to find $\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?$ when w = 300 m.

The area of a rectangle is denoted by Area = length * width.

Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:

• $A=2w^2$

Differentiate this equation with respect to time t.  

• $\displaystyle \frac{dA}{dt} =4w \cdot \frac{dw}{dt}$

Let's plug known values into the equation:

• $\displaystyle \frac{dA}{dt} =4(300) \cdot (24)$

Simplify this equation.

• $\displaystyle \frac{dA}{dt} =1200 \cdot 24$
• $\displaystyle \frac{dA}{dt} =28800$

The area is changing at a rate of 28,800 m²/yr at this point in time.