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 34 m per year at a time when the region is 260 m wide. How fast is the area changing at that point in time

Answer 1

Answer:

The rate of change of area is 35360 m /year

Step-by-step explanation:

Let the width of rectangle be w

We are given that  a rectangle that is twice as long as it is wide

So, Length of Rectangle  = 2w

Area of rectangle A = Length \times Breadth = $w \times 2w= 2w^2$

Differentiating area with respect to time

$A=2w^2\\\frac{dA}{dt}=4w\frac{dw}{dt}$

We are given that $\frac{dw}{dt} = 34m/year$ and w = 260

$\frac{dA}{dt}=4(260)(34)=35360$

Hence the rate of change of area is 35360

Answer 2

Using implicit differentiation, it is found that the area is changing at a rate of 35360 m²/year.

The area of a rectangle of length l and width w is given by:

$A = lw$

Applying implicit differentiation, we the rate of change is given by:

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

In this problem:

• 260m wide, thus $w = 260$.
• Twice as long as it is wide, thus $l = 2w = 520$.

Also:

$\frac{dl}{dt} = 2\frac{dw}{dt}$

Since $\frac{dw}{dt} = 34, \frac{dl}{dt} = 68$

Then:

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

$\frac{dA}{dt} = 520(34) + 260(68) = 35360$

The area is changing at a rate of 35360 m²/year.

A similar problem is given at https://brainly.com/question/2194008