The length of a rectangle is increasing at a rate of 3 cm/s and its width is increasing at a rate of 9 cm/s. When the length is 10 cm and the width is 5 cm, how fast is the area of the rectangle increasing?

Question

anonymous · Answer

Well, you want to use $$
A = l\times w
$$

anonymous · Answer

And then you differentiate.

anonymous · Answer

I would like to see it step by step if possible

anonymous · Answer

same idea as before
Area of the rectangle is $A=xy$
taking the derivative, and using the product rule, you get $$A'=xy'+yx'$$ 
plug in the numbers your are given

anonymous · Answer

$$
\frac{dA}{dt} = \frac{dl}{dt}w + l\frac{dw}{dt}
$$

anonymous · Answer

We are given that $dl/dt = 3\text{cm/s}$ and $dw/dt = 9\text{cm/s}$

anonymous · Answer

Substituting that in, we get: $$
\frac{dA}{dt} = 3w\ \text{cm/s} + 9l\ \text{cm/s}
$$

anonymous · Answer

They also give us $l$ and $w$, so we have to substitute them in too.

anonymous · Answer

what about just being in cm

anonymous · Answer

never mind, i was trying to figure the wrong thing out that was in the question itself