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Calculus/Two towers cable optimization

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Question
Two vertical towers of heights 60ft, and 80 ft stand on level ground, with there bases 100ft apart. A cable that is stretched from the top of one pole to some point on the ground between the poles, and then to the top of the other pole. What is the minimum possible length of the cable required?

Answer
Let the 60 ft tower stand on the left and the 80 ft tower stand on the right.  Let the distance of the point the cable is pegged to the ground be x from the 80-feet tower.

Then the distance from the peg to the 60-feet tower is 100-x.

The slanted side to the lower 60-feet tower (by Pythagoras theorem) is
         sqrt (60^2 + (100-x)^2  )  = sqrt (13600 - 200x   + x^2 )

The slanted side to the higher 80-feet tower (by Pythagoras theorem) is
         sqrt (80^2 + x^2  )     = sqrt (6400 + x^2)

Hence the total length of cable required is

  D =  sqrt (13600 - 200x   + x^2 )   +   sqrt (6400 + x^2)

To minimize D with respect to x, we need to find the critical point of x as x vary between 0 and 100.  Then prove that the critical point in fact makes D a minimum.

To find the critical point of D, find dD/dx set it to zero then solve for x.
i.e.          ( 2x - 200 )/sqrt (13600 - 200x   + x^2 ) + 2x /sqrt (6400 + x^2) = 0

squaring both sides (and manipulating) yield
    (x-100)^2 (6400+x^2)   = x^2 (13600-200x+x^2)
The algebra is very involved, but I got
      (7x-400)(x-400) = 0

that will give us x = 400/7    or x = 400 (inadmissible because x is between 0 and 100).

To prove minimum, one can
1) use the first derivative test for D  for x between 0 and 100.
2)  use the second derivative test and show that d'(400/7) is positive.
3)  Plot D for x between 0 and 100 and show that the curve is truly a minimum when x = 400/7.  

Calculus

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Amos

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I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

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