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
+118703 
