How fast is the area of rectangle increasing? 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?
So on internet I found a solution but I didn't do that way and I am still thinking that I am not wrong but the answer is not the same. I am gonna write both the solutions which I found on int and by myself and I will be waiting your help.
Which I found on the int: 
$A=lw$  then take derivative $\frac{dA}{dt}= \frac{dl}{dt}.w + l.\frac{dw}{dt}$
using given number $\frac{dA}{dt}= (8)(10) + (20)(3)$
My answer: given numbers--> $\frac{dl}{dt}= 8$,   $\frac{dw}{dt}=3$,  $l=20$, $w =10$
so $A=wl$  when I wanna write $w$ in terms of $l$ ----> $l=2w$
so $A=2w*w$ when I take derivative of it ---> $\frac{dA}{dw}= 4w $
according to chain rule $\frac{dA}{dt}= \frac{dA}{dw}\frac{dw}{dt}$ 
when I put the numbers ----> $4w*3$ and we know that $w=10$ 
It should be 120. I think I found my mistake but still couldn't understand why. I write $w$ in terms of l but if I do the other way then the result is 160. What am I doing wrong?
 A: One millisecond later, the sides are $20.008$ and $10.003$ and the relation $l=2w$ is no more true.
The rate of increase of the area must be close to
$$\frac{20.008\cdot10.003-20\cdot10}{10^{-3}}=140.024.$$
With one microsecond, we get
$$\frac{20.000008\cdot10.000003-20\cdot10}{10^{-6}}=140.000024.$$
This confirms the answer $140$.
The reason why your method doesn't work is because
$$\frac{20}{10}\ne\frac{8}{3}.$$
A: While it is true that at this moment in time $A = 2w^2 = 200$
The length and width are not changing uniformly.
If they were then it would be correct to say $A = 4w \frac {dw}{dt}$
But as they are changing at different rates, you need to use the chain rule.
Perhaps a visualization will help.

We have the rectangle at time $t$ and at time $t+1$ and and the red and green rectangles are the approximate change at some intermediate time.
the green areas sum to $l (dw)$ and red areas $w (dl)$
A: You need to think of $l$ and $w$ as functions of time.
$l(t) = L + 8t$ where $L$ is the initial length.  And $w(t) = W+3t$ and, yes, $\frac {dl}{dt}=8$ and $\frac {dw}{dt} = 3$ but $l \ne 20$ and $w\ne 10$.  (That'd mean they are constant functions.  They aren't.)  $l(t_0) = 20$ and $w(t_0) = 10$ at time $t_0$.
To make things simple we can let $t_0=0$ and $l(t) = 20 + 8t$ and $w(t) = 10 + 3t$ and $l(0) =20$ and $w(0) = 10$.
Now we just can't say $l(t) = 2w(t)$ because that just isn't true.  You could say $ {l(t)} = \frac {20+8t}{10+3t}{w(t)}$ and $A(t) =  \frac {20+8t}{10+3t}{w(t)}^2$ but... that just makes things complicated.  (If I had more time and energy I'd be curious to try to figure that and see if it is the same but.... I don't.)
