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The length of a rectangle is increasing at a rate of 4 inches per second while its width is decreasing at a rate of 3 inches per second. At what rate, in square inches per second, is the area of the rectangle changing when its length is 23 inches and its width is 18 inches?

 Nov 2, 2021
 #1
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The area is increasing at a rate of 6 in^2/sec.

 Nov 2, 2021
 #2
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Would you mind explaining how you got it? I'm not sure if it is right.

Guest Nov 2, 2021
 #3
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The area is decreasing, not increasing!!!

 

23 * 18 = 414

27 * 15 = 405

31 * 12 = 372

35 * 9 = 315

39 * 6 = 234

43 * 3 =129

Guest Nov 3, 2021
 #4
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The length of a rectangle is increasing at a rate of 4 inches per second while its width is decreasing at a rate of 3 inches per second. At what rate, in square inches per second, is the area of the rectangle changing when its length is 23 inches and its width is 18 inches?

 

We are not given a start point but I only care about the rates so where it starts, I mean at what size, doesn't matter.

So I am going to say that at  time 0 the length is 23inches  and the width=18 inches

When t=0   L=23,  w=18

 

\(\frac{dw}{dt}=3\\ so\quad w=3t+18\\ and\\ \frac{dl}{dt}=4\\ so \quad l=4t+23\\ A=wl \\ A= (3t+18)(4t+23)\\ A=12t^2+(69+72)t+(18*23)\\ A=12t^2+141t+(18*23)\\ \frac{dA}{dt}=24t+141\\ when \;\;t=0\\ \frac{dA}{dt}=141\;\;inch/sec\)

 

 

 

 

 

LaTex:

\frac{dw}{dt}=3\\
so\quad w=3t+18\\
and\\
\frac{dl}{dt}=4\\
so \quad l=4t+23\\
A=wl \\
A= (3t+18)(4t+23)\\
A=12t^2+(69+72)t+(18*23)\\
A=12t^2+141t+(18*23)\\
\frac{dA}{dt}=24t+141\\
when \;\;t=0\\
\frac{dA}{dt}=141\;\;inch/sec

 Nov 3, 2021
 #5
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Thank you so much for explaining!

Guest Nov 6, 2021
 #6
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You are welcome :)

Melody  Nov 7, 2021

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