Related Rates - Rectangles

Question

A rectangle has a length of 1 inches and a width of 12 inches whose sides are changing. The length is decreasing by 2 in/sec and the width is shrinking at 5 in/sec. What is the rate of change of the area?

Answer 1

Let x = width in inches at time t sec, y = length at time t and A = area at time t

We know that dx/dt = -5 and dydt = -2

Find dA/dt when x = 12 and y = 1:

A = xy

So, dA/dt = x(dy/dt) + y(dx/dt) = 12(-2) + 1(-5) = -29 in²/sec

The area is decreasing at the rate of 29 square inches per second.

Answer 2

Estefani, this looks like a generalized product rule problem (unless you are asked to find how the area is changing at a given time when x = 12 and y = 1).

Area = (x)(y)

deriv(Area) = x'y + xy'

deriv (Area)= (-5)(y) + (x)(-2)

and that's all, unless you are asked to find the value at a given time. If the question is asking for the rate of change is for x = 12 and y = 1, then plug those values in.

Hope this helps, but let me know--Andrew