TITLE: Minimizing fuel usage for small boat between given points
QUESTION [9 upvotes]: I'm 12 years old and my I was getting bored this afternoon so my dad gave me this math problem (he said it was supposed to be hard, and that I should do some research to learn how to solve it).
"A small boat moving at $V$ km/h uses fuel at a rate given by the function $$q = 8 + \frac {V^2}{50}$$  where q is measured in litres/hour. Determine the speed of the boat at which the amount of fuel used for any given journey is the least."
I had no idea how to do it until i found some stuff on the internet about "calculus". I figured out that i might have to work out a formula for the total fuel consumed (fuel rate multiplied by time). But when i tried to do this, I found that I have created another variable (distance) when i was trying to write V in terms of d/t.
I am really stuck, i feel like i have worked out how to do these types of problems, but this particular one I cannot solve. 
I dunno, would you guys maybe be able to solve it, or is it too high-level, maybe i should take it to my math teacher?

REPLY [2 votes]: For a 12 year old, a bit of experimentation may be easier.
Try looking at the extremes first. Choose easy numbers, such as V=1 and V=100.
For V=1, q=8+$\frac{1}{50}$. In one hour, you'll burn 8.02 liters of fuel to travel 1 kilometer. The biggest part is the constant "8"
For V=100, q=8+$\frac{10000}{50}$ In one hour, you'll now burn 208 liters of fuel, but you traveled 100 kilometer. That's better, only 2.08 liter per kilometer. The biggest part is now $\frac{V^2}{50}$ 
Let's experiment further. What if both parts would be equal? What if $8 = \frac{V^2}{50}$ ? That means $V^2=400$ or V=20. In that case, q=16. In one hour, you travel 20 kilometer, for only 16 liters. That's only 0.80 liter per kilometer!
Ok, so we see that if we go slowly, the 8 part is important, and if we go fast, the $V^2$ part is important, and in the middle you have a better solution. But what is best?
The trick we use in math is to find the best speed V so that going faster (V+dV) uses more fuel, and going slower (V-dV) also uses more fuel. dV is a very small number. In fact, we'll make it so small that $dV^2$ is small enough to ignore. Thus, $(V+dV)^2 = V^2 + 2*V*dV$. 
So that means we'd have a $q+dq = 8 + \frac {V^2 + 2*V*dV}{50}$. That's a bit unruly. Let's simplify it by looking around V=20. $16+dq = 8 + \frac {400 + 40*dV}{50}$ or $dq = \frac {4}{5}*dV$. Now that's nicer. If I go slightly faster than 20 km/h, my fuel consumption per hour goes up just as fast. And if I go slightly slower, my fuel consumption goes down equally fast.
But that means the fuel consumption per kilometer doesn't change around V=20! Now 20 is a somewhat lucky choice of V. I could do the same for V=10, and then I would see that $q = 10, dq = \frac {2}{5} dV$. My fuel burn rate goes up slower than my speed, at only 40%. So it's better to go faster than 10 kilometer per hour.
So, if I hadn't guessed V correctly from the trick $8=\frac{v^2}{50}$, I would have had to solve dq/dV = q/V. That's a bit of writing but you'll get V=20 all the same.