Calculus - Trig Maximum Value Problem When the rules of hockey were developed, Canada did not use the metric system. Thus, the distance between the goal posts was designated to be six feet. If Sidney Crosby is on the goal line, three feet outside one of the goal posts, how far should he go out (perpendicular to the goal line) to maximize the angle in which he can shoot at the goal? 
(Hint: Determine the values of x that minimize θ)
I have tried to solve this question by using tan to find an equation for θ for the small triangle and the large triangle, but am getting nowhere. 
 A: The target is actually $3$-dimensional.  I did not realize they played hockey in Flatland. (In very casual games we did have a rule about not raising the puck. Otherwise, nobody would agree to be goalie.) 
We will need a common picture to refer to, and I don't want to draw it, so need to describe it. Draw a horizontal line segment $AB$ of width $6$, with $A$ on the right. Let $P$ be Crosby's current location, $3$ feet to the left of $A$. Let $Q$ be a position he may go to, with $QP$ perpendicular to $AB$. We want to maximize the angle $AQB$. Let us explore what happens if we make $QP=x$. Note that
$$\angle AQB=\arctan(9/x)-\arctan(3/x).$$
(Alternately, we could use $\text{arccot}(x/9)-\text{arccot}(x/3)$. It is a matter of taste. If we look up the derivative of $\text{arccot}$, the algebra is simpler because the Chain Rule calculation is more pleasant.)
Remark: Instead of maximizing $\angle AQB$, you may wish to maximize its tangent. Note that $\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}$. That will make the algebra a lot nicer. 
