Calculus Max Min Distance Problems
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Max Min Distance Problems

 

Example:

 
 Person 1 is standing at some point.
 Person 2 is standing 20 miles to the east of Person 1 ...
 

 

 
 At the same moment, Person 1 starts walking south at 4 miles per hour,
 and Person 2 starts walking west at 2 miles per hour ...
 

 

 
 How long after they start walking, will the distance between them be a minimum?
 
 OK, the first thing to do is to figure out the equation
  for the distance between them.
 We drop a coordinate axis on them (I hope it didn't hurt much).
 It will make things easier if we put the origin at the place person 1 started ...
 

 

 
 So by arranging the axis this way, each person walks along one of the axis. 
 That is, for person 1 position, only the Y coordinate changes
 for person 2, only the X coordinate changes.
 You will see in a second that that makes the math easier.
 
 Person 1 is moving south along the Y axis.
 When person 1 starts walking, they are at (0,0).
 At any time (T) after that, they are at:
 

 (0, (0-4T))

 
 Person 2 starts out at a point 20 miles east of person 1. That is, at (20,0)
 Person 2 walks west at 2 miles per hour.
 That means at any point in time they are at:
 

 ((20 - 2T), 0)

 
 If you look at the diagram above, 
 you can see that the distance between them is the hypotenuse of a triangle.
 That means we can use the old Pythagorean distance formula!
 So at any time T, the distance between them is:
 

 

 
 Multiplying this out ...
 

 

 
 Combining terms ...
 

 

 
 We want to find where the distance, D, is a minimum.
 That means we need to check the point where the slope is zero and the endpoints.
 One endpoint is where time = 0, that is, the starting point.
 There really isn't another end point. Time can run forever here.
 That means we need to calculate a limit as time approaches infinity.
 To get the slope equals zero point,
 we need the first derivative of the distance equation.
 
 I find it easier to change roots to fractional exponents when doing this.
 A square root is the same as raising the expression to the one half power.
 So ...
 

 

 
 Don't forget that this is a power chain rule problem,
 so we have to deal with an "inside part."
 
 We need to find the point where D'=0,
 so it would help to write this as a real fraction...

 

 

 
 When the numerator equals zero, the whole expression will equal zero.
 As long as the denominator does not equal zero.
 So ...

  0 = 40T - 80

 T = 2 hours

 
 Check to make sure the denominator is not equal to zero at T = 2.
 We only need to find the value under the square root radical.
 

 20T2 - 80T + 400

 20(2)2 - 80(2) + 400

 80 - 160 + 400

 320

 
 OK, the denominator is not equal to zero.
 Now lets check the second derivative to make sure that this is a minimum...
 Use the quotient rule, and hold on ...
 

 

 
 Do some simplifying ...
 

 

 
 Solve this mess for when T = 2 ...
 

 

 
 The second derivative is positive.
 That means the T=2 hours point is a minimum.
 
 So what is the distance between the two people at that point?
 The distance formula was ...
 

 

 
 Plug in T = 2 and chug away.
 You should get ...

 D = 17.89 miles

 
 The endpoints are where T = 0 and where T approaches infinity.
 At T = 0, the first two terms under the radical equal zero 
 so the distance is the square root of 400, so

 

 D = 20

 
 Where T approaches infinity,
  the value will be controlled by the highest exponent term under the radical.
 It takes a bit of fancy math to prove this one, 
 but it is pretty obvious that the distance approaches infinity.
 
 So the minimum distance of 17.89 miles happens when T = 2.
 

   copyright 2005 Bruce Kirkpatrick

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