Continuously changing Dimensions of a Rectangle 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?

"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
$$
\frac{dA}{dt}\bigg\lvert_{t=3} 
    = L(3)\frac{dW}{dt}\bigg\lvert_{t=3} + W(3)\frac{dL}{dt}\bigg\lvert_{t=3}
    = 14\cdot 3 + 29(-2) = -16.$$
