Anonymous asked in Science & MathematicsMathematics · 3 weeks ago

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

  • 3 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!

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