Expert: Paul Klarreich - 11/9/2010

Question
I need a setup for this: It didn't come to me so here goes the question. While doing the question, I got only 2 times. If it decides to travel around the shore, time ~23.56 min. while if it goes through water on a straight line, time = 30 min.

The driver of an amphibious vehicle wishes to cross a circular lake 4 miles in diameter to a point on the shore diametrically opposite. Vehicle's speed on land = 16mph and in water = 8mph. It can go around the lake on the shore, or directly across through the water, or in a straight line through the water to an intermediate point on the shore and then along the shore the rest of the way. Find min/max times required for trip. You must express time in terms of angle between diameter and path of the vehicle through the water.  

Answer
Questioner:   Phillip Larsen
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Calculus Optimization Problems

Question:  I need a setup for this: It didn't come to me so here goes the question. While doing the question, I got only 2 times. If it decides to travel around the shore, time ~23.56 min. while if it goes through water on a straight line, time = 30 min.

>> That might well be the case -- sometimes the min/max occurs at an endpoint.

A driver wishes to cross a circular lake 4 miles in diameter to a point on the shore diametrically opposite. Vehicle's speed on land = 16 mph and in water = 8 mph. It can go around the lake on the shore, or directly across through the water, or in a straight line through the water to an intermediate point on the shore and then along the shore the rest of the way. Find min/max times required for trip.

You must express time in terms of angle between diameter and path of the vehicle through the water.

>> I think I would rather express it differently.  See pic.  Using  t for theta,

We cross water:  segment  AP
We cross land:   arc PB

We get AP by law of cosines:

AP^2 = 2^2 + 2^2 - 2(2)(2) cos (pi - 2t)

But

cos (pi-2t) = cos pi cos 2t + sin pi sin 2t

 = - cos 2t

AP^2 = 8 + 8 cos 2t

AP = sqrt(8 + 8 cos 2t)

AP = 2 sqrt(2 + 2 cos 2t)
and time needed = 8*AP = 16 sqrt(2 + 2 cos 2t)

We get arc PB = central angle * radius

arc PB = 2t * 2 = 4t
and time needed = 16*4t = 64t

Total time = 16 sqrt(2 + 2 cos 2t) + 64t

That will be a bit messy, but not impossible.