# 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?

• Why isn't it $8*3 = 24$. If the area is $lw$ then the area after a second will be $(8l)(3w) = 24wl$ and so on. am I missing something? – Yanko Jan 17 at 20:59
• $l$ is twice $w$ only at that particular time and it's not a general functional relationship between these two values. – Matteo Jan 17 at 21:01
• Your expression for $\frac{dA}{dt}$ is correct. It should be $140$. – John Douma Jan 17 at 21:03
• @Yanko, $8$ and $3$ are rates of increase of each size of the rectangle. So your expression is not correct. The rate of increase of the area is correctly $$\frac{dA}{dt} = w\frac{dl}{dt} + l\frac{dw}{dt}$$. – Matteo Jan 17 at 21:10
• @Yanko after one second the area is $(l+8)(w+3)$. Not $(8l)(3w)$. – fleablood Jan 17 at 21:22

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

$$\frac{20}{10}\ne\frac{8}{3}.$$

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 $$w (dl)$$

• thank you! I think I understand now but I can't open the link, I think it's kind of bug. Do you mind uploading the picture another website. – Disintegrators Jan 17 at 21:55
• More precisely, the rates shouldn't be equal but proportional to the respective sides. – Yves Daoust Jan 17 at 22:02
• @Disintegrators I thought I had. Looks fine on my screen. – Doug M Jan 17 at 22:35

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.)