Optimization and related rates trig.

[amcelroy13]

Homework Statement

A woman at point A on the shore of a circular lake with radius 2 miles wants to arrive at point C diametrically opposite A on the shore of the lake in the shortest time possible. She can walk at 4 mph and row a boat at 2 mph. To what point on the shore of the lake should she row before walking o minimize time? Verify your answer.

Homework Equations

Distance = velocity x time
t=d/v
total time = d in water/2mph + d on land/4mph

The Attempt at a Solution

I know that I must use angles to find the point B but I have no idea exactly what to do. If i could find the relation I could easily optimize it, but I really have no idea how to do it.

 

[LCKurtz]

Your equation: total time = d in water/2mph + d on land/4mph is a good start. Draw a picture showing the person traveling a little way around the circumference to a point B. Call the center of the lake O. Call angle ACB $\theta$ and notice that angle AOB is $2\theta$. You should be able to get the walking distance arc AB and the rowing distance BC in terms of $\theta$ to put in your equation for time. Then you are on your way.

 

[amcelroy13]

so if arc AB = r*2theta (i think) would BC = 2/sin(pi-2theta)?

 

[amcelroy13]

never mind, that was dumb, its not a right triangle, BC would be r*(pi-2theta)

 

[amcelroy13]

as of now i have gotten:

time=((r(pi-2theta)/2)+r*2theta/4
r=2 so i have:
time = pi-(pi*theta)+2pi
I have no idea how to optimize... any help?

 

[LCKurtz]

You don't have the equation for BC correct. You are going to need some trig functions. In your figure, drop a line from B perpendicular to the diameter AC and call the intersection P. Now CBP is a right triangle and the rowing part BC is its hypotenuse. You can get OP from the little triangle OPB with a trig function and CP = OP + r. Now you can get the rowing distance CB from triangle CPB with another trig function. You have the walking arc length correct.

 

[amcelroy13]

I figured it out, thanks for the help