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