The length of a rectangle is increasing at a rate of 8 cm/s and its width is increasing at a rate of 5 cm/s. when the length is 13 cm and the width is 11 cm, how fast is the area of the rectangle increasing?

A = xy
A' = xy' +x'y
A' = (13 cm)*(5 cm/s) +(8 cm/s)*(11 cm)
.. = (65 +88) cm^2/s
.. = 153 cm^2/s . . . . . . . the rate at which rectangle area is changing


Check:
.005 seconds before the time of interest, the area was
.. (13 cm -.005*8 cm) * (11 cm -.005*5 cm) = 12.96*10.975 = 142.236 cm^2
.005 seconds after the time of interest, the area is
.. 13.04* 11.025 = 143.766 cm^2
So the average rate of change over that 0.01 second interval is
.. (143.766 -142.236)/(.01) = 153 cm^2/s . . . . close enough

B.-2,3 because it is touch the -3 i think that the answer