This is problem 49 page 219:
A man is in a boat 2 miles from the nearest
point on the coast. He is to go to a point
Q, located 3 miles down the coast and 1
mile inland. If he can row at 2 miles per
hour and walk at 4 miles per hour, toward
what point on the coast should he row in
order to reach point Q in the least amount
of time.
Need a picture:
Find x to minimize time
2 miles
3-x
x
1 mile
Q
3 miles
Use the Pythagorean theorem twice to
find the distance to row and walk
To row:
x 4
To walk:
(3  x )  1
2
2

The time to row is:  hour  x 2  4 miles
 2 miles 
Notice I used unit analysis to find I had to
divide by 2 and not multiply by 2
The time to walk is:
 hour 


 4 miles 

( 3  x )  1miles
2


The primary equation is the total time:
T 
1
2
x 4
2
1
(3  x )  1
2
4
We don’t need a “secondary equation”. Why?
The primary equation only has one
variable, there is no need to substitute
for a second variable.
To minimize, find the derivative = 0
1
T 
x 4
2
2
1
(3  x )  1
2
4
This will be good chain rule practice
1 2
1
T '
x 4

22

Outside

1
2
 1 1
2 x    4  (3  x ) 2  1

2

Inside
Outside

1
2

( 2 ( 3  x )(  1)) 

Inside
Do you see another
chain rule here?
We need to solve this mess = 0
1 2
1
T '
x 4

22


 1 1
2 x    4  (3  x ) 2  1

2

1
2

1
2

( 2 ( 3  x )(  1)) 

First, let’s clean it up a bit by multiplying constants

2

1
T '
x 4

2

1
2

2


1
 x    4  (3  x )  1



1
2
( 3  x ) 

Wow, let’s cheat and use MathPert to graph
this and find the solution
Click on this
Now this
Select “Type it in”
note the use of ( ) everywhere and
the use of negative exponents
now
click OK
Make sure this is the equation we want
and it is. Click OK again
Here is our graph, we want where it equals 0,
ie, the x-intercept
click this about 4 times to get a better
view
This looks like
the point
Use the Point and Slope Tool
click it and center the crosshairs
on the intercept
y = 0 and x = 1
this is where the derivative
is equal to 0!
We just use a ‘numerical” technique to solve the equation
Just to make sure, evaluate at x = 1:

2

1
T '
x 4

2

1

2
 x    1 4  ( 3  x ) 2  1



1
2

(3  x ) 

You should get:
1
2 5

2
0
4 5
The man should aim for a point 1 mile
down the coast

2
T '  1  x  4
2

1

2
 x    1 4  ( 3  x ) 2  1



1
2
( 3  x ) 

To review what I typed into MathPert:
1/2((x^2+4)^(-1/2)(x))-1/4((3-x)^2+1)(-1/2)(3-x))
“Count” the number of opening (( versus the
number of closing )), they must match in any
expression. I could also type, which is easier:
x/2sqrt(x^2+4)-(3-x)/4sqrt((3-x)^2+1)