For my homework, I was given this brainteaser:
You’re sunbathing on the island shown on the map below. The island is six miles from shore at
the closest point, and the nearest store is a convenience store seven miles down the beach. If
you can row at 4 miles an hour, and you can walk at 5 miles an hour, where should you land to
get a bag of tortilla chips in the least possible time? (Ignore tides, currents, and sharks. No fair
renting a helicopter.)
I found the equation for the time it takes to get to the store to be:
$\frac{x}{4} + \frac{7-\sqrt{x^2 – 36}}{5}$
I then found the derivative:
$\frac{1}{4} – \frac{2x}{10 \sqrt{x^2-36}}$
I didn't find any zeroes for the derivative inside the domain of the function [6, $\sqrt{85}]$, so I answered that the shortest time would be achieved by rowing straight to the store.
The teacher said the answer was eight. Where did I make a mistake?
edit: Given image: 
edit: Some closure:
My email to him:
On the brainteaser, the store is only 7 miles away, how can the answer be 8
😛For my formula, 10 gave the same answer as 8 in the formula the answer uses.
i.e., rowing 10 miles lands you 8 miles from the point across from the
island. Of course, sqrt(85) is less than 10, so it's only faster if you can
then walk 7 – 8 = -1 miles to the store and turn back time a little.
His response:
It assumes you can go back in time.
Best Answer
Putting the origin on the left, your travel time is
$d(x)=\cfrac{1}{4} \sqrt{6^2+x^2} +\cfrac{1}{5}(7 - x) $
The derivative is
$d'(x) = \cfrac{1}{4} \cfrac{x}{\sqrt{6^2+x^2}} -\cfrac{1}{5} $
$d'$ is negative on $(-\infty,8)$ so $d$ is strictly decreasing on $[0,7]\subset(-\infty,8)$. Since we want the minimal value of $d$ on $[0,7]$, we'll get it at $x=7$, $f(7)=\cfrac{\sqrt{85}}{4}$.
I'm not sure about your second term but your $x/4$ means your $x$ is on the vertical axe... So it'd be like doing it this way: