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?

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Answer 1 · 2016-12-01T21:57:36

#140" cm"^2"/"s#

Let us set up the following variables:

# {(l, "Length of Rectangle (cm)"), (w, "Width of Rectangle (cm)"), (A, "Area of Rectangle ("cm^2")"), (t, "Time (s)") :} #

We are told that:

#(dl)/dt = 8# cm/s (const), and, #(dw)/dt = 3# cm/s (const)

The Area of the rectangle is:

# A=lw #

Differentiating wrt #t# (using the product rule) we get;

# (dA)/dt = (l)((dw)/dt) + ((dl)/dt)(w) #
# :. (dA)/dt = 3l + 8w #

So when #l=20# and #w=10 => #
# \ \ \ \ \ (dA)/dt = 3*20 + 8*10 #
# :. (dA)/dt = 60 + 80 #
# :. (dA)/dt = 140 " cm"^2"/"s#