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

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SUMMARY

The discussion focuses on calculating the rate of change of the area of a rectangle with a fixed area of 75 cm², where the length is three times the width and the width changes at a rate of 2 cm/second. The correct approach involves using the formula A = lw, leading to A = 3w². The derivative of the area, A', is calculated using the related rates formula, yielding A' = 60 cm²/sec. The final answer confirms that the rate of change of the area is 60 cm²/sec when the width is 5 cm.

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a78
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Homework Statement


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.

Homework Equations

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

Homework Statement

Homework Equations

The Attempt at a Solution

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

a78 said:

Homework Statement


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.

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##.

Homework Equations

The Attempt at a Solution


A=75 A'=?
L=3x= 15 L'=6
W=x=5 W'=2

Again, the derivative of constants are zero.

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

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.
 
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