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Extreme Value Problem - Analytic
Subcase 1: The point Q is on the shoreline (d2=0).

Problem setting: A man with a boat at point S at sea wants to get to point Q on the shore. Point S is distance d1 from the closest point P on the shore. The points P and T are at a distance of d from each other.

Question: If the man rows with a speed of v_r and walks with a speed of v_w at what point R should he beach the boat in order to get from point S to point Q in the least possible time? .

This problem is solvable using calculus without any kind of calculator and it points out several interesting aspects of extreme values.

 

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1st Sample Problem - Straightforward Setting
A person on a boat in a lake is 4 km from the shore and must go to a point 12 km down the shoreline. The person can row the boat at 2 km/h and can walk at 5 km/h. To what point on the shore should he row the boat in order to reach his destination in the shortest possible time?

Solution:
Let's make a sketch of the problem.

 

1. Let x=\overline {PR} . Then, 12-x=\overline {RQ} .
Using Pythagoras' theorem \overline {SR} = \sqrt {x^2 + 16} so that time spent rowing is: \frac{{\overline {SR} }}{v_r}= \frac{{\sqrt {x^2 + 16} }}{2}
2. \overline {RQ} = 12 - x so that time spent walking is: \frac{{\overline {RQ} }}{v_w} = \frac{{12 - x}}{5}
It follows that the Total Time function is: t(x) = \frac{{\sqrt {x^2 + 16} }}{2} + \frac{{12 - x}}{5}
This function needs to be minimized, so we take the derivative of t(x) with respect to x , set it equal to 0 and solve for x .
3. Here is the derivative: t'(x) = \frac{x}{{2\sqrt {x^2 + 16} }} - \frac{1}{5} .
     Notice that the distance ' d= 12 km ' is gone from this derivative!
4. Putting t'(x)=0 , we get: \frac{x}{{2\sqrt {x^2 + 16} }} = \frac{1}{5} or 5x = 2\sqrt {x^2 + 16}
Squaring both sides of this equation we get: 25x^2 = 4 (x^2 + 16)
Solving for the positive value of x : 21x^2 - 64 = 0 or x=1.75 .

Answer: The man should beach the boat 1.75 km down the shore.
     Notice that this solution does not depend on ' d= 12 km'. In the simulation, change the value of d and notice that the location of the minimum of t does not change.

 

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2nd Sample Problem - General Problem
a) Find the total time t as a function of x=\overline {PR} using d1 , d , v_r and v_w as parameters (constants).
b) Find a formula for x for the minimum value of t .
c) Check that this general solution works for the parameters of the first problem.
d) Discuss when this minimum value exists and why?
e) What is the role of the distance " d km down the shoreline" in your answer? Why?

Solution:
Let's make a sketch of the problem.

 

a) Let x=\overline {PR} .
1. Using Pythagoras' theorem: \overline {SR} = \sqrt {x^2 + d1^2} so that time spent rowing is: t_r=\frac{{\overline {SR} }}{v_r}= \frac{{\sqrt {x^2 + d1^2} }}{v_r}
2. \overline {RQ} = d - x so that time spent walking is: t_w= \frac{{\overline {RQ} }}{v_w} = \frac{{d - x}}{v_w}
It follows that the Total Time function is: t(x) = \frac{{\sqrt {x^2 + d1^2} }}{v_r} + \frac{{d - x}}{v_w}
b) This function needs to be minimized, so we take the derivative of t(x) with respect to x , set it equal to 0 and solve for x .
Here is the derivative: t'(x) = \frac{x}{{v_r \cdot \sqrt {x^2 + d1^2 } }} - \frac{1}{{v_w }} .
Putting t'(x)=0 , we get: \frac{x}{{v_r \cdot \sqrt {x^2 + d1^2 } }} = \frac{1}{{v_w }} or v_w \cdot x = v_r \cdot \sqrt {x^2 + d1^2 }
Squaring both sides of this equation and solving, we get:

 

    v_w^{\,2} \cdot x^2 = v_r^{\,2} \cdot (x^2 + d1^2 )
   (v_w^{\,2} - v_r^{\,2} )x^2 = v_r^{\,2} \cdot d1^2
    x = \frac{{v_r \cdot d1}}{{\sqrt {(v_w^{\,2} - v_r^{\,2} )} }}

 

Answer: The man should beach the boat \frac{{v_r \cdot d1}}{{\sqrt {(v_w^{\,2} - v_r^{\,2} )} }} km down the shore.

 

c) Check: The parameters of the first problem are d1=4 , v_r=2 and v_w=5 .
Substituting, we get: x = \frac{{2 \cdot 4}}{{\sqrt {(5^2 - 2^2 )} }} = \frac{8}{\sqrt {21}}=1.75 .
Since this was our answer, the formula checks out.

 

d) A minimum exists only if {(v_w^{\,2} - v_r^{\,2} )}>0 , that is only if the man walks faster than he rows.
This is because if he rows faster than he walks - then he should row the whole way!

 

e) Notice that the distance d = \overline {PQ} is gone from the derivative of t!
This means that the minimum is independent of d - that is, d has no role in the answer.
Why? This is a bit tricky to answer ...  It is because the value of x , that is the position of R, is the point at which the relative differences in rates of rowing and walking and the distance d1 to the shore 'balances out' and d has no effect on this.


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Page last modified on January 25, 2009, at 12:41 AM