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### Where to Land on the Beach?

```
Date: 5/22/96 at 16:26:29
From: Anonymous
Subject: Calculus

An island is at point A, 6 miles offshore from the nearest point B on
a straight beach.  A store is at point C, 7 miles down the beach from
B. If a person can row at the rate of 4 mi/hr and walk at a rate of
5 mi/hr, where should the person land in order to go from the island
to the store in the least possible time?
```

```
Date: 5/31/96 at 16:58:8
From: Doctor Anthony
Subject: Re: Calculus

With the problem as given, the best course of action is to row
straight from A to C.  If you land on the beach between B and C you
will take longer.  I suspect that the numbers in the question should
be changed to give a bigger advantage to walking on the beach, but I
will work through the calculus anyway to show the method.

Let x be the distance from B alng the beach where the person lands.
The distance he rows is then sqrt(x^2 + 36) and the rowing time is
sqrt(x^2+36)/4.  The distance he walks will be (7-x), and time walking
is (7-x)/5.

Total time = sqrt(x^2+36)/4 + (7-x)/5

Now differentiate with respect to x.

dt/dx = (1/4)(1/2)(1/sqrt(x^2+36))(2x) - (1/5) = 0 for max. or min.

(x/4)(1/sqrt(x^2+36)) = (1/5)

5x = 4sqrt(x^2+36)  Square the equation
25x^2 = 16(x^2+36)
25x^2 = 16x^2 + 576
9x^2 = 576    And taking square roots
3x = 24
x = 8

And we see that this means landing beyond point C to obtain a
theoretical minimum turning point.  In practice we should row directly
to C.  I feel sure the numbers given in the question are not those
intended, but the method to use is exactly as outlined above.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 08/07/2003 at 05:06:01
From: Ajit
Subject: Calculus/Maxima

In the above problem, Dr. Anthony says that in real life you would row
directly to point C. Why is your calculus at variance with the obvious
answer of going directly to to point C?

5x = 4sqrt(x^2+36)  Square the equation

Now look at this equation carefully. Coefficients 4 and 5 are really
the speed of rowing and walking respectively. Correct?
Consider the situation where the two speeds are equal. Would you then
get (Speed of walking)x = (speed of rowing)sqrt(x^2+36) and hence
x = sqrt(x^2+36) which is clearly unacceptable?

This is precisely why with speeds as given viz. 4 mph and 5 mph Dr.
Anthony has arrived at x = 8 miles. The whole analysis seems to be erroneous.

A cursory glance at the situation tells us that:
If the speed of rowing >> speed of walking then one should land close
to B to minimize the time taken.
If speed of rowing << speed of walking then go close to C.
If the two speeds are equal then go directly to C.
In any other situation the landing point must be between B and C.
Your analysis does not reflect this at all. Why?
```

```
Date: 08/07/2003 at 12:58:16
From: Doctor Peterson
Subject: Re: Calculus/Maxima

Hi, Ajit.

Let's deal with two issues: first, why the calculus yields a result
that has to be overridden; and second, why your "cursory" expectations
are wrong.

When calculus is applied to an optimization problem, we have to
consider not only the equations, but also constraints. In this case,
it was assumed when the equations were written that the landing point
was on the segment BC, so that x<7. If this were not done, then the
equation would have needed an absolute value:

Total time = sqrt(x^2+36)/4 + |7-x|/5

This was not done, in order to simplify the work, since it is obvious
that x must be between 0 and 7 for a valid solution. But it means
that the constraint on the equation has to be checked.

When it was found that the optimal value of x was 8, outside the
bounds of the equation, this meant that the true optimum had to be
found at a boundary point. If the absolute value formulation had been
used (which in effect combines this equation, with its constraint,
with another equation that is applied in a different region, to make
a piecewise-defined function), then it would have been found that
neither function has a minimum, and again the boundary between the
functions would have to be checked. The true optimum would be not at
a place where the derivative is zero, but where the derivative does
not exist.

(You'll see below what happens when the two speeds are the same,
which causes a slightly different problem.)

>A cursory glance at the situation tells us that:
>If the speed of rowing >> speed of walking then one should land close
>to B to minimize the time taken.
>If speed of rowing << speed of walking then go close to C.
>If the two speeds are equal then go directly to C.
>In any other situation the landing point must be between B and C.
>Your analysis does not reflect this at all. Why?

Sometimes what is "obvious" is true; sometimes not.

Your claims don't really make sense to me; perhaps you wrote it
wrong. If I row fast, I would expect to spend all or most of my time
rowing, so I would land near C; if I walk faster, I would want to do
more walking, by landing closer to B. I would never expect to land at
B exactly, since just a small amount of extra rowing would save a
good bit of walking. And I wouldn't have to do both at the same speed
in order to make a direct route to C a good idea. I would expect that
there would be some minimum ratio of rowing to walking speed that
would make C the best landing point.

Let's see under what conditions the landing point is actually between
B and C. I'll replace the speeds 4 and 5 with u and v, and the
distances 6 and 7 with a and b; then we have

t = sqrt(x^2+a^2)/u + (b-x)/v

The derivative is then

dt/dx = (1/u)(1/2)(1/sqrt(x^2+a^2))(2x) - (1/v)

and when we set this to zero we get

x = ua/sqrt(v^2 - u^2)

If u >= v, this is undefined. In that case, we have to compare the
times when x=0 and when x=b; these are

t(0) = a/u + b/v
t(b) = sqrt(b^2 + a^2)/u

It can be shown that if u >= v, then t(0) > t(b), so the optimal
strategy is to go directly to C. That agrees with my version of your
ideas.

The other extreme will be when the optimal x is >= b, as in our
original problem. Then again the optimal strategy will be to go to C.
When does this happen? When

ua/sqrt(v^2 - u^2) >= b

This can be solved to give

u/v >= b/sqrt(a^2 + b^2)

That is, when the ratio of rowing to walking speeds is at least that
of the maximal walking distance to the maximal rowing distance, it
will be better to row all the way. (Note that the right side here is
less than 1, so that this case includes the case where u >= v.)

So it turns out that the optimal landing point will be between B and
C when

u/v < b/sqrt(a^2 + b^2)

And it will NEVER be best to land at B (unless, of course, b=0 so
that B=C; or it a or u is 0, which don't make sense).

I may well have made some mistakes here or there, but since the
result agrees with my intuition, it "must" be correct ;-)

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/07/2003 at 22:05:11
From: Ajit
Subject: Thank you (Calculus/Maxima)

Thanks a lot Dr Peterson. I obviously got confused between
points B and C in my "thoughts". But now you have cleared
my doubts and thanks are due to you and the Math Forum.
```
Associated Topics:
High School Calculus

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