Perhaps you can do the drawing. Let $L$ be the location of the light. (It is on the ground!) Let $F$ be the location of the man's feet, and let $H$ be the top of his head. Let $B$ be the bottom of the building. Draw the horizontal line $LB$. Draw the vertical line $FH$ to represent the short thin man.
Draw the line through $L$ and $H$. Let it meet the wall of the building at $S$. Then $BS$ is the length of the shadow.
It is convenient to call the distance $LF$ by the name $x$. We are told the man is walking at $3$ feet per second, so $\frac{dx}{dt}=3$.
Let $y=BS$. Note that triangles $LBS$ and $LFH$ are similar, so
$$\frac{y}{30}=\frac{4}{x}.$$
We may want to rewrite this as
$$y=\frac{120}{x}.$$
Differentiate. We get
$$\frac{dy}{dt}=-\frac{120}{x^2}\frac{dx}{dt}.$$
Now it is almost over.
Not quite. You missed a key word in the problem. The first sentence tells you the cylinder is decreasing in height, but with a constant volume. If something is constant, then it is not changing. If it is not changing, its rate of change (i.e. derivative) is zero.
But wait, how can the cylinder be decreasing in height, yet remain constant in volume? This implies that the radius must be increasing, to balance out the lost volume from a decreasing height.
You are correct in differentiating $V=\pi r^2 h$ with respect to time. It's just implicit differentiation. So we have
$$\frac{dV}{dt} = \pi r^2 \frac{dh}{dt} + 2\pi r h \frac{dr}{dt}.$$
However, at this point we must recognize that the problem just told us the volume is constant! If something is constant, it doesn't change, so its derivative is zero. That is,
$$0 = \pi r^2 \frac{dh}{dt} + 2\pi r h \frac{dr}{dt}.$$
Plugging in the given rate $dh/dt$, and evaluating at $r=3$ inches, and $h=4$ inches, we have
$$0 = \pi (3 \text{ in})^2 \left(\frac{-1 \text{ in}}{5 \text{ sec}}\right) + 2\pi (3 \text{ in}) (4 \text{ in}) \frac{dr}{dt}$$
$$0 = -\frac{9}{5}\pi\ \frac{\text{in}^3}{\text{sec}} + 24\pi \ \text{in}^2\frac{dr}{dt}.$$
$$\frac{9}{5}\pi\ \frac{\text{in}^3}{\text{sec}} = 24\pi \text{in}^2\frac{dr}{dt}$$
Now we divide by $24\pi\ \text{in}^2$. Notice how I deliberately involve the units in my calculations. It's not necessary, but it makes our work easier to check. For example, if the end result is not in the units we expect, then we know we make an error in our algebra.
$$\frac{9}{120}\frac{\text{in}}{\text{sec}} = \frac{dr}{dt}.$$
And yes, our radius is indeed increasing, with the correct units.
Best Answer
(I realize this is a little late, but hopefully someone else can find this helpful!)
Your idea about similar triangles was exactly right. Having an accurate diagram makes things a lot easier! Solving for $x$, we have (assuming $y=30-16t^2$): $$\begin{align} \frac{y}{10+x}&=\frac{6}{x} \\[0.25ex] yx&=60+6x \\[0.5ex] x(y-6)&=60 \\ x&=\frac{60}{y-6} \\ x&=\frac{60}{24-16t^2} \\ x&=\frac{15}{6-4t^2} \end{align}$$
All we need to do now is find $\frac{dx}{dt}$ and evaluate it at $t=1$, which isn't terribly difficult. $$\begin{align} \frac{dx}{dt}&=\frac{-15(-8t)}{(6-4t^2)^2} \\ \frac{dx}{dt}&=\frac{120t}{2^2(3-2t^2)^2} \\ \frac{dx}{dt}&=\frac{30t}{(3-2t^2)^2} \\ \left.\frac{dx}{dt}\right|_{t=1}&=\frac{30}{(3-2)^2} \\[0.75ex] &=30\:\mathrm{ft/s} \end{align}$$