Help me with my calculus problem . Please be hurry !?

A boat is 4 miles from the nearest point on a straight shoreline ( call it point A )
Then a restaurant is on the shoreline , and 6 miles away from A .
A woman plans to row to appoint on shore , and then walk to the restaurant . If she can walk at 3 mph , at what speed must she be able to row so that the quickest way to get to the restaurant is to row directly .

http://i1023.photobucket.com/albums/...rg25/photo.jpg
this is the picture

2. So think of x as the point that he lands on shore where A is at the origin and B is at x = 6. So the distance he would have to walk on the shore would be 6-x and the distance in the boat would be $\sqrt{4^{2}\; -\; x^{2}}$.
You know that she has a walking speed of 3 mph and we want to know v rowing mph then the total time of her journey is shown as
$](\sqrt{4^{2}\; -\; x^{2}})/v + (6-x)/3$
where t = distance/velocity.

Derive the above expression and set to zero to find the lowest amount of time needed to travel. (Answer should be in terms of x and v and you should derive with respect to x)

You should get something like this:
$dt\; =\; \frac{1}{2v}\left( \frac{2x}{\sqrt{4^{2}\; -\; x^{2}}} \right)\; -\; \frac{1}{3}\; dx = 0$

You then know she is looking for the rowing speed such that her lowest time will be to row straight to the beach meaning she should land at the beach at x = 6 and you can directly enter that into the equation above and solve for v.

$dt\; =\; \frac{1}{2v}\left( \frac{2\cdot 6}{\sqrt{4^{2}\; -\; 6^{2}}} \right)\; -\; \frac{1}{3}\; dx\; =\; 0$

and v = $\frac{\sqrt{52}}{8}\; mph$

3. hi I think you made a mistake , it must be sqrt(x^2+4^2) not sqrt(x^2-4^2) Thank anyway .