[Math] Minimizing fuel usage for small boat between given points

Question

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?

Answer 1

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 $V=20$.