グローバルビジネスで役立つ数学（14）フェリーの速度を求める（英語版）

Hi, there!

If my memory is correct, I had a graduation school trip in my second
grade at high school. We went to Kyushu on a ferry, and it was a me-
morable voyage for me, for I had one mission to complete upon arri-
val.

I had a correspondence with a girl in Kyushu then, and I was schedu-
led to meet her for the first time in the precincts of a specific temple.
The students were not allowed to act on their own out there, but my
homeroom teacher gave me a “get-out-of-jail-free card,” given a kind
of love story in my adolescence.

The teacher lived in my neighborhood, and I guess he had no choice
but to oblige my wish for the neighbors’ sake.

I was trying to keep my excitement to myself; however, several friends
realized I had a secret plan to do at our destination. Finally, they sneak-
ingly followed me up to an appointed meeting place.

By all means, I managed to spend about thirty minutes or so talking
with her. Still, it was a nagging regret that I should have met with her
in a shrine, not in a temple because I might have been able to hide my
excitement if I fully “enshrined feelings” deep in my heart.

Here comes the word problem today.

When a ferry is sailing bound for Kyushu, the fuel cost is proportional
to the square of its speed relative to water. Besides that, there are fixed
costs, which do not depend on the speed and are equal to $p$ ($/hour).
At what velocity will the total cost per mile be the lowest? Don’t forget
to use a coefficient of proportionality $(k)$ to find the answer $(v)$.

In retrospect of the voyage in the school trip above, let me find the ans-
wer as described below. Here is my solution to the problem.

The variable part of expenses hinges on the speed like
$q = kv^{2}$, where $k$ is a coefficient of proportionality.

The total cost per hour is expressed by the formula
$C = p + q = p + kv^{2}$.

In one hour, the ferry travels a distance equal to $v$.
Therefore, the cost per mile is given by

$C = C(v) = \frac{C}{v} = \frac{p + kv^{2}}{v} = \frac{p}{v} + kv$

The resulting expression is a function of the speed $v$.
Let’s check out its extreme values.

$C(v) = \frac{C}{v} = \frac{p + kv^{2}}{v} = \frac{p}{v} + kv$
$C'(v) = 0$, then $(\frac{p}{v} + kv)’ = -\frac{p}{v^{2}} + k = \frac{kv^{2} – p}{v^{2}}$

$\frac{kv^{2} – p}{v^{2}} = 0$, then $\left\{\begin{array}{l}kv^{2} – p = 0\\
v^{2} \neq 0\end{array}\right.$

therefore, $v = \sqrt{\frac{p}{k}}$

The real voyage of discovery consists not in seeking new landscapes,
but in having new eyes.

Stay tuned, and expect to see my next post.

Keep well.

Frank Yoshida

￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣￣
【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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