Is the optimal solution to this problem to row straight to the store? 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
  :P
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.

 A: 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:

A: For fun, we do the problem the hard way.  Forget about the location of the store for now, say it is far  away.
We can row to the nearest point on the shore, then walk. Imagine setting out instead at an angle $\theta$ to the shortest line to the shore. Then we row an extra distance of $6\sec\theta -6$, but we walk less by $6\tan\theta$, so the amount of time "saved" over the direct to shore way is
$$\frac{6\tan\theta}{5} -\frac{6\sec\theta-6}{4}.$$
Now we could use calculus, the differentiation is in some ways more pleasant than if we use $x$'s. 
But let's see what can be done without calculus. Change to sines and cosines. After a little manipulation we arrive at
$$\frac{6}{\cos\theta} \left(\frac{\sin \theta}{5} -\frac{1-\cos\theta}{4}\right).$$
Using half-angle identities, and letting $\phi=\theta/2$, we find that the time saved is
$$\frac{12\sin\phi}{\cos\theta}\left(\frac{\cos\phi}{5}-\frac{\sin\phi}{4}\right).$$
We save time if the above expression is positive, that is if $\tan \phi\lt \frac{4}{5}$. Thus increasing $\phi$ is always a good idea until $\tan\phi=\frac{4}{5}$.  This corresponds to $\tan\theta=\frac{40}{9}$. 
But $\tan\theta=\frac{40}{9}$ is $\frac{240}{9}$ miles down the beach, a lot more than $7$ miles.  So the best strategy is to row directly to the store.  (The whole calculation is probably unnecessary. The speeds $4$ and $5$ are close to each other, as are the distances $6$ and $7$. So the diagonal is the best way.)
A: Let $y$ be the landing position relative to the island (so the store is at $y=7$).  Then the distance from the island to the landing position is $\sqrt{36+y^2}$, and the walking distance to the store from there is $|y-7|$.  The total time is
$$
T(y)=\frac{1}{4}\sqrt{36+y^2}+\frac{1}{5}|y-7|,$$
and the derivative with respect to $y$ is
$$
T'(y)=\frac{y}{4\sqrt{36+y^2}}+\frac{1}{5}\text{sgn}({y-7})$$
for $y\neq 7$; the function $T(y)$ is not differentiable at $y=7$, but has the value $\frac{1}{4}\sqrt{85}$ there.  It remains to check for any other local extrema.  There are two cases: $y>7$ and $y<7$.  Clearly $T'(y)$ is positive for all $y>7$, since both terms are positive in that case.  For $y<7$, we need to check for solutions to
$$
\frac{y}{4\sqrt{36+y^2}}-\frac{1}{5}=0.$$
The only solution, $y=8$, is not in the range where this form of $T(y)$ applies.  We conclude that there is no local minimum of $T(y)$ except the one at $y=7$, and so you are correct that the fastest way to reach the store is by rowing straight to it.
