Question

One side of a rectangle increases 2cm/s, while the other side decreases at 3cm/s. How fast is the area of the rectangle changing when the first side equals 20 cm and the second side equals 50 cm ?

Related Rates problem

Answer 1

The area of the rectangle is given by:\[\bf A=L \times W\]Let 'L' b the first side and 'W' be the second side. Let dL/dt represent the change in the first side length and dW/dt represent the change in the second side length with respect to time. Differentiating both sides of the Area equation we obtain:\[\bf \frac{ dA }{ dt }=\frac{ dL }{ dt }W+\frac{ dW }{ d t}L\]Now we just plug in the given values for each:\[\bf \frac{ dA }{ dt }=2(50)+3(20)=100+60=160 \ cm^2/s\]
@muffin

Answer 2

okay that makes sense, for some reason the answer on the sheet is 40cm^2/s ..?

Answer 3

100-60 gives 40

Answer 4

muffin has it. There is a sign error on the derivative in the second term.

Answer 5

okay