Minimizing fuel usage for small boat between given points 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? 
 A: If you can assume that $V$ is constant through the journey you need to make the quantity you are interested in - the amount of fuel - one of the variables. If $t$ is the time in hours then $qt=Q$ is the amount of fuel, which you want to reduce as much as possible. Note also that $Vt=D$ will be the distance travelled in km, and "for any given journey" suggests that this should be taken as constant for the journey in question. You need to find which constant $V$ minimises the use of fuel $Q$ for a journey of length $D$.
So first multiply the equation by $t$ to get $$Q=8t+\frac {DV}{80}$$
Now you can control $V$ and are not really interested in $t$ so get rid of $t=\frac DV$ so that $$Q=\frac {8D}V+\frac {DV}{80}$$
After these manipulations you now have $Q$, the quantity you want to minimise, expressed in terms of the constant $D$ and the quantity you can control, namely $V$.
When faced with a problem like this, the first task is to transform it into a form which is expressed in terms of the quantities you are interested in, and eliminating others.
I'll leave you to do the interesting job of finding the final answer from this form of equation.
A: 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.
A: Consider the journey is $L$ km long and lasts $t$ hours. What is the relation between $L$, $t$ and $V$ ?

 By definition $V=L/t$.

Now compute how many liters of fuel will be consumed during the journey ? Call this number $F$. 

 $F$ is the consumption rate multiplied by time, so $F=qt$. 

Now comes the difficult part. Here's a little help : compute the number of liters of fuel consumed per km (call this number $G$ for instance) and express it so that it depends on $V$ and not on $t$ or $L$. 

 $G=F/L=qt/L$ Since $V=L/t$, you have $t=L/V$, therefore $G=q/V=8/V+V/50$.

We now have the number of liters of fuel consumed per km that depends only on $V$. We wish to make this number as small as possible. I do not know your background. There is a technique call differentiation (computation of derivative) that gives a response to this problem, but you may not know it. 
With a plotting program, you can draw the curve of $G$ as a function of $V$
where $V$ can take values from $0$ to a large number, say $100$ for instance.
If you don't have it, place your mouse over the area under this paragraph, you'll see how it looks like.

 

You notice that there is a minimum at a value that you can read on the axis.

 The value of the minimum is located at $V=20$ liters per hour.

How can one find the value without a graphic method ? Compute $\frac52G$ and try to write the formula as simple as you can. The answer is quite symmetric.

 You should find $\frac52G=\frac{20}V+\frac{V}{20}$. 

Now here is a little reasoning. Replace $V$ in such a way that the equation for $\frac52G$ is simply $x+\frac1x$. Knowing that this function has a unique minimum, you should be able to prove the result you have found graphically. 

 Setting $x=\frac{V}{20}$ we have the formula $\frac52G=x+\frac1x$. The minimum of $G$ is the same as the minimum of $\frac52G$ and then it is the minimum of $x+\frac1x$. Notice that if you replace $x$ by $\frac1x$, this does not change the value: This means that the minimum is obtained for $x$ and for $\frac1x$. But there is a unique minimum, which means that $x=\frac1x$. The only solution is therefore $x=1$ and then $V=20$. 

