The dimension of a rectangle are continousl changing. The width increases at a rate of 3 in/s while the length decreases at the rate of 2 in/s. At one instant the rectangle is a 20-in square. How fast is its area changing 3 seconds later?
The answer is -16. $$\begin{align}A &= LW,\, 20 = LW \\[1ex] \frac{dA}{dt} &= L\frac{dW}{dt} + W\frac{dL}{dt}\\[1ex] \frac{dA}{dt} &= \frac{20}W\cdot 3 + W\cdot(-2)\\[2ex] t=3\implies W &= (3\text{ in/s})*3\text s = 9\text { in}\\[1ex] \frac{dA}{dt} &= \frac{20}9\cdot3 + 9\cdot(-2) = \frac{34}3\end{align}$$
Is my assumption at t=3 wrong? any hint?
$\endgroup$2 Answers
$\begingroup$Your expression for dA/dt is correct. All you need to do is substitute the correct values. In this case, we know that L=W=20 at t=0. At t=3 them, L=20-3*2 and W=20+3*3. dW/dt and dL/dt were given. If you substitute these values in your equation for dA/dt, you will get the expected answer. Note that your value for W at t=3 is not correct; you forgot to take into account the value at t=0.
$\endgroup$ 2 $\begingroup$"The rectangle is a 20-in square" means that it is a $20\times 20$ square; that is, $L(0) = W(0) = 20$. Three seconds later, $L(3) = 20-3\cdot 2 = 14$ and $W(3) = 20 + 3\cdot 3 = 29$, so that $$ \frac{dA}{dt}\bigg\lvert_{t=3} = L(3)\frac{dW}{dt}\bigg\lvert_{t=3} + W(3)\frac{dL}{dt}\bigg\lvert_{t=3} = 14\cdot 3 + 29(-2) = -16.$$
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