Hello, rawkstar!

This one takes a bit of work.

Hope you're sitting down . . .

$\displaystyle \text{A man with a boat is located at point }P\text{ on the shore}$

. . $\displaystyle \text{of a circular lake of radius 5 miles.}$

$\displaystyle \text{He wants to reach the point }Q\text{ diametrically opposite }P\text{ as soon as possible.}$

$\displaystyle \text{He plans to paddle his boat at an angle }\theta\text{ to }PQ\:(0 \le \theta \le \frac{\pi}{2}) $

. . $\displaystyle \text{to some point }R\text{ on the shore, then walk along the shore to }Q .$

$\displaystyle \text{If he can paddle at 2.9 mph and walk at 4 mph,}$

. . $\displaystyle \text{ what is the shortest possible time it will take him to reach }Q?$

Code:

* * *
* * R
* o
* x * * *
* *5 y
* * @ * *
P o - - - - o - - - - o Q
* 5 O *
* *
* *
* *
* * *

The center of the circle is $\displaystyle \,O.$

. . And: .$\displaystyle OP = OQ = OR = 5$

Let $\displaystyle \theta = \angle RPQ$

. . Note that: .$\displaystyle \angle ROQ = 2\theta$

Let $\displaystyle \,x \,=\,\overline{PR}$, the distance he rows.

Let $\displaystyle y \,=\,\text{arc}(RQ)$, the distance he walks.

Find the distance $\displaystyle \,x.$

In isosceles triangle $\displaystyle POQ,\:\angle POQ \,=\,\pi - 2\theta$

Use the Law of Cosines:

. . $\displaystyle x^2 \:=\:5^2 + 5^2 - 2(5)(5)\cos(\pi - 2\theta) \;=\;50 + 50\cos2\theta $

. . $\displaystyle x^2 \;=\;50(1 + \cos2\theta) \;=\;100\left(\dfrac{1+\cos2\theta}{2}\right) \;=\;100\cos^2\!\theta $

. . $\displaystyle x \;=\;10\cos\theta$

He rowed $\displaystyle 10\cos\theta$ miles at 2.9 mph.

. . This took: .$\displaystyle T_1 \:=\:\dfrac{10\cos\theta}{2.9}$ hours.

Find the distance $\displaystyle \,y.$

The arc length $\displaystyle RQ$ is: .$\displaystyle \text{radius} \times \text{angle}$

Hence: .$\displaystyle y \:=\:5(2\theta) \;=\:10\theta$

He walked $\displaystyle 10\theta$ miles at 4 mph.

. . This took: .$\displaystyle T_2 \:=\:\dfrac{10\theta}{4} \:=\:\dfrac{5}{2}\theta$ hours.

The total time is: .$\displaystyle T \;=\;\dfrac{10}{2.9}\cos\theta + \dfrac{5}{2}\theta$

And *that* is the function you must minimize.

*Go for it!*