I remember entering a design contest at school where we had to make a self propelled vehicle out of toilet rolls, elastic, etc... During that process one of the Physics teachers mentioned that to maximize the range of a projectile one should aim at 45 degrees. This is another story of how a teacher was wrong. It is also the story of how he was right.
So, where does this result come from? Well, I had learned how to derive it in my first year at Uni (I never actually did Physics before that). The initial conditions are that the projectile is launched at a point (let's say the origin) at some initial speed $$ v=\left \langle v_x,v_y \right \rangle $$. The final location is thus given by physics as:
Where $$ g=-9.79 $$ in Cape Town (thanks Physics 1004W) and $$ v_y = \sqrt{\left | v \right |^2 - v_x^2} $$
You can then solve the system for an equation in $$ v_x $$ and maximize $$ x_f $$. I thought that was quite cool, and time passed, and after my 4th year of studying I interned with Oracle. One day we were out at lunch at a golf course where people were hitting these golf balls from a balcony. So there was some height involved. We somehow got onto that topic about 45 degrees being optimal and I remember wondering what the optimal angle was. I worked it out napkin style so that the system now read:
but there is a fair amount of work involved in solving this. So, here I am nearly a year later, and just as I'm about to fall asleep the problem comes creeping back. I somehow realized that one can use that strange thing we learned in 2nd year called "Lagrange multipliers". In this case you want to maximize $$ f(v_x,t) = v_x t $$ subject to the constraint $$ g(v_x,t) = v_y t + \frac{1}{2}g t^2 = -h $$. We write:
The first equation can be solved for lambda which can be thrown into the second to obtain:
$$ v_x^2 = \sqrt{\left | v \right |^2 - v_x^2}(\sqrt{\left | v \right |^2 - v_x^2}+gt) $$
We can then get $$ t $$ out of the 3rd equation since it's a quadratic and throw that into this to arrive at:
$$ v_x^2 = -\sqrt{\left | v \right |^2 - v_x^2} \sqrt{\left | v \right |^2 - v_x^2-2gh} $$
This is a quadratic in $$ v_x^2 $$ representing negative of positive direction. We know that $$ v_x $$ is positive. Similar for things like $$ t $$ by the way. Anyway, we solve for $$ v_x $$ now and obtain:
$$ v_x = \sqrt{\frac{\left | v \right |^2(\left | v \right |^2-2gh)}{2(\left | v \right |^2-gh)}} $$
You can find the angle as $$ \arctan(\frac{v_y}{v_x}) $$ if you like but note that its typically smaller than 45 degrees. Something that really sort of struck me initially was that this depends on the initial speed, $$ \left | v \right | $$. I guess it sort of makes sense in hindsight though.
Anyway, there you have it. Lagrange multipliers are still useful for lunch time chat :)
The next thing to look at would be if you let $$ h $$ vary. So you could have $$ h $$ be some hill or something. Then $$ h(x) = h(v_x t) $$ and the partial derivatives are nice because you get something times $$ h'(v_x t) $$ in each case. Roughly speaking, I think you can actually do this in general by solving for $$ t $$ in equation 3 using $$ h^{-1}(h(v_x t)) $$ and dividing by $$ v_x $$. You will end up with something a bit nasty though.
You may rather want to consider a numerical scheme or perhaps an evolutionary algorithm like this which I wrote to evolutionarily solve the Brachistochrone problem... but that's a story for another day.