# I need help on calc problem?

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 39 m per year at a time when the region is 480 m wide. How fast is the area changing at that point in time?

### 2 Answers

- stanschimLv 73 weeks ago
A = lw = (2w)(w) = 2w^2

Differentiating with respect to t gives:

A'(t) = 4ww'

A'(t) = 4(480)(39) = 74880 square meters per year

- 3 weeks ago
Assume the growth is constrained so that the length is always twice the width: l = 2w

The zone is rectangular, so A = lw = (2w)w = 2w^2

We want to know the rate of change of the area over time, so take derivatives with respect to time on both sides: A' = 4w * w'. The w' on the right comes from chain rule.

We're given a width and a fixed rate of change of the width in the problem, so we can plug those into this equation to get the rate of change of the area:

A' = 4(480 m)(39 m/year) = 74880 m^2/year

Hope this helps!