# What is the rate of change of the area of a rectangle?

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1. May 27, 2017

### a78

1. The problem statement, all variables and given/known data
Find the rate of change of the area of a rectangle whose area is 75cm^2. The length is 3 times the width. The rate of change of the width is 2cm/second.

2. Relevant equations

3. The attempt at a solution
A=75 A'=?
L=3x= 15 L'=6
W=x=5 W'=2

A'=L'W+LW'
A'= (6)(5)+ (15)(2)
A'=60cm/sec

Last edited: May 27, 2017
2. May 27, 2017

### BvU

And it looks like homework, so use the template and show your attempt!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

3. May 27, 2017

### BvU

Much better ! The problem statement helps you to formulate what you actually need to solve.
easiest if you express the rate of change as a $d\over dt$, as you did.
Do not forget to keep track of the units: an area is cm2, so a growth rate for an area can not be cm/s !!!
Other than that, you are doing quite well !

4. May 27, 2017

### LCKurtz

While you seem to understand how to work this problem, your writeup could be much better. I will intersperse comments below.

A rectangle whose area is $75$ has constant area, so that isn't what you mean. What you mean is a rectangle has length 3 times its width, so its area is $A = lw =3w^2$. The rate of change of $w$ is $2$, and you want the rate of change of $A$ when its area is $75$.

Again, the derivative of constants are zero.

Since you already have the correct answer, I am going suggest a better way to write it up. Since you have $A = 3w^2$ you know that $A' = 6ww'$, where $' = \frac d {dt}$. That is called the related rate equation -- everything is a function of $t$ so the derivatives aren't $0$. At the instant when $A = 75$, you have figured out that $w = 5$ and $w'$ is given as $2$. Just put those numbers in the related rate equation and you have your answer.