Optimization / Related Rate Searchlight I can't seem to figure out how to set up this problem. I'm a bit confused on the relationship between the variables here; I've seen a similar problem involving a searchlight but the angle of $\theta$ at which the beam leaves the light was given. Without that I'm not sure how to manipulate the information given.
The problem reads: A searchlight is 100m from the nearest point on a straight highway. As it rotates the searchlight casts a horizontal beam that intersects the highway in a point. If the light revolves at a rate of pi/6 rad/s, find the rate at which the beam sweeps along the highway as a function of $\theta$. For what value of $\theta$ is this rate minimized?
I don't need the problem solved for me, I can do that, I just don't know how to set this up. The relationship is confusing to me. Thanks.
 A: Draw a picture. The road might as well be the $x$-axis, and the searchlight could be at $(0,100)$. 
Label the position of the serchlight $S$. The nearest point on the road to $S$ might as well be called $O$. At a certain time $t$, the searchlight illuminates the point $X=X(t)$ on the road. Put $X$ say to the right of $O$. Call the $x$-coordinate of $X$ by the name $x=x(t)$. 
Let $\theta=\theta(t)$ be the angle the beam has travelled through in time $t$. We measure angle with respect to the line $ON$. 
For the picture, let $\theta=\angle XON$. Then $\tan\theta=\dfrac{x}{100}$.
Over to you! We know $\frac{d\theta}{dt}$ and want $\frac{dx}{dt}$. 
Added: We could now solve for $x$ in terms of $\theta$. I prefer to differentiate immediately. Using the Chain Rule, we get
$$\sec^2\theta \frac{d\theta}{dt}=\frac{1}{100}\frac{dx}{dt}.$$
On the assumption the rotation is counterclockwise, we get that $\frac{d\theta}{dt}=\frac{\pi}{6}$. That gives
$$\frac{dx}{dt}=\frac{100\pi}{6}\sec^2\theta.$$
The minimum value of $\sec^2\theta$ is reached at $\theta=0$ (and $2n\pi$ for any positive integer $n$). As the intuition readily verifies, the minimum rate of change of $x$ is when $x=0$, that is, when the beam illuminates the nearest point $N$ on the road.
