# Minimum problem involving rates and basic trig

• JOhnJDC

## Homework Statement

A man at point A on the shore of a circular lake of radius 1mi wants to reach the opposite point C as soon as possible. He can walk 6mi/h and row his boat 3mi/h. At what angle theta to the diameter AC should he row?

## The Attempt at a Solution

My diagram looks like this: I first drew the diameter of the circle AC with A being on the left. Next, I called the point to which the man should row B and I drew a line from A to B (B being somewhere between theta = 0 and theta = pi/2). Finally, I drew another line from O (the center of the lake) to B. Therefore, angle CAB = theta and angle COB = 2theta. Moreover, AC=2, so OB=1, AO=1, and OC=1.

My initial equation for the shortest total time that it will take the man to reach point C is:

T = AB/3 + 2theta/6

Now, I know that I need to calculate the derivative of T and set it equal to zero to minimize T. However, I'm not sure how to calculate AB.

By the law of cosines:

(AB)2=AO2 + OB2 -2ab(cosAOB), which equals:

(AB)2=12 + 12 - 2cos(AB) or 2-2cos(AOB)

This is where I get confused. I set cos(AOB) equal to cos(2theta - 2pi), but that isn't working out for me. The correct answer has AB = 2cos(theta). I've been working at this for a while and I don't understand how to get AB = 2cos(theta). I would greatly appreciate your help.

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I figured this out, if anyone is interested. I set AOB equal to cos(pi-2theta) instead of cos(theta - 2pi).

(AB)2 = 2 - 2cos(pi - 2theta)
(AB)2=2-2(cos(pi)cos(2theta) + sin(pi)(2sin(theta)cos(theta))
Because cos(pi)=-1 and sin(pi)=0, we have
(AB)2 = 2-2(-cos(2theta))
(AB)2=2(1+cos(2theta))
(AB)2=2(2cos2(theta)
(AB)2=4cos2(theta)
AB=sqrt[4cos2(theta)]
AB=2cos(theta)

Therefore, the shortest total time is

T=(2cos(theta))/3 + (2theta)/6

T'=[-2sin(theta) +1]/3
Setting T' equal to zero yields sin(theta)=1/2

Therefore, theta equals pi/6.

However, T''=[-2cos(theta)]/3, so the critical point sin(theta)=1/2 is a maximum. Is this right? Shouldn't it be a minimum?

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