# The Length Of A Rectangle Is Increasing At A Rate Of 8 Cm/s

Related rates problem:

The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 3 cm/s.  When the length is 20 cm and the width is 10 cm, how fast is the area of the rectangle increasing?

We are interested in calculating the rate of change of the area. This can be interpreted as the derivative of the area with respect to time.

The rate of increase of its length is:

dl/dt = Vl = 8 cm/s

The rate of increase of its width is:

dw/dt = Vw = 3 cm/s

The area of the rectangle is

A = length * width = l * w

Deriving the area with respect to time, we have:

dA/dt = l(dw/dt) + w(dl/dt)

Substituting the above values, we have:

dA/dt = (20)(3) + (10)(8) = 140 cm2/s

Then the rate of change of the area, when the length is 20 cm and the width is 10 cm, is 140 cm2/s

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Let `a(t)` and `b(t)` be functions denoting length and width of the rectangle at time `t.` Since area of rectangle is product of length and width we have

`A(t)=a(t)b(t)`

To get the the rate of change we need to calculate first derivative of function `A` (velocity is derivative of displacement over time). Therefore, by using product rule we get

`A'(t)=a'(t)b(t)+a(t)b'(t)`

Now we are interested to find `A'(t_1)` where `t_1` is time at which `a(t_1)=20` and `b(t_1)=10.` And since rates of change are constant (equal for all t) for both length and width we have `a'(t_1)=8` and `b'(t_1)=3.`

Hence,

`A'(t_1)=8cdot10+20cdot3=140`

When the length is 20 cm and the width is 10 cm, the area of the rectangle is increasing at `140cm^2//s.`

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