Question: 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?
So I know for this problem that there is a third quantity that I can assign a variable to and solve an equation for. However, after giving this problem a great deal of thought, I can confidently say I am stuck. Where can I go from here?
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$\begingroup$Let $x$ be the length and $y$ the width. Then: $$E=xy$$ $$E'=x'y+xy'$$ We have from the given information that $x'=4$, $y'=-3$ (because it's decreasing), $x=23$ and $y=18$.
Can you continue?
$\endgroup$ 10 $\begingroup$Let $x$=length, $y$=width.
$$A=xy$$
You know $\frac{dx}{dt}=4 \;\frac{\text{inches}}{\text{sec}}$, $\frac{dy}{dt}=-3 \; \frac{\text{inches}}{\text{sec}}$.
$$\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}$$
$$\frac{dA}{dt}=23\cdot {-3}+18\cdot{4}$$
$$\frac{dA}{dt}= 3 \frac{\text{inches}^2}{\text{sec}}$$
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