APPLIED OPTIMIZATION problem?

Question

A man is floating in a rowboat one mile from the (straight) shoreline of a large lake. A town is located on the shoreline one mile from the point on the shoreline closest to the man. He intends to row in a straight line to some point P on the shoreline and then walk the remaining distance to the town. To what point should he row in order to reach his destination in the least time if:

a) he can walk 5 mph and row 3 mph? (miles per hour)=mph

b) he can walk 5 mph and row 4 mph?

please explain the steps

thanks

Answer 1

he should row and not walk

Answer 2

Label A the closest point on the shore and the B the point where the town is. Then his initial distance from A is 1 mile and the distance form A to B is 1 mile. He needs to row straight line to the point C which is between A and B. Mark AC as x. x has a range from 0 to 1 mile. Then his total time is:
case a: t = sqrt(x^2 + 1)/3 + (1 - x)/5. This function has a minimum at x = 3/4 mile. So he should row to the point 1/4 of a mile from the town and then walk 1/4 of a mile.
case b: t = sqrt(x^2 + 1)/4 + (1 - x)/5. This function within the x range has a minimum for x = 1. Therefore he should row straight line to the town.

Answer 3

Let T be the point representing the town, M be the man and P be the point closest to the man on the shore. Triangle MPT is a right triangle with MP=1 and PT=1.

Let X be the point between P and T which is x miles from P. If the man rows towards that point, he will row sqrt(x²+1) miles and then walk (1-x) miles to the town. The traveling time is then

t(x) = sqrt(x²+1)/r + (1-x)/w, where r and w are the rowing and walking speeds, respectively. Differentiating, we get...

dt/dx = x/[r*sqrt(x²+1)] - 1/w. We want this to be zero in order to minimize t(x), so

x/[r*sqrt(x²+1)] = 1/w
wx = r*sqrt(x²+1)
w²x² = r²(x²+1)
w²x² = r²x² + r²
x² = r²/(w²-r²)
x = r/sqrt(w²-r²)

Plug in w=5 and r=3 or 4 and you get the answers x=3/4 for question a, and 4/3 for question b.

I thought I did something wrong, because for question b, I got x=4/3 which means X is PAST the town. This is outside the domain of the t(x) function that I created beause the land distance for x>1 is no longer (1-x), rather it is (x-1). This means we don't have a local minimum for 0<=x<=1. Rather, we can show that dt/dx < 0 for 0<=x<=1 in this case, which means t(x) is decreasing on this whole interval and its minimum will therefore be at x=1. More generally, if r>=w/sqrt(2), there will be no x in [0,1] satisfying dt/dx=0 and the quickest way to the town is to row straight at it.

So. After all this long-windedness...

a) Go for x=3/4, and walk the remaining 1/4 of a mile.
b) Row straight towards the town.