another derivative word problem

[PaulKraemer]

Hi,

I am stuck on the following problem:

Two vertical poles of lengths 6 ft and 8 ft stand on level ground, with their bases 10 ft apart. Approximate the minimal length of cable that can reach from the top of one pole to some point on the ground between the poles, and then to the top of the other pole.

If 'd' is the horizontal distance from the 6 ft pole to the point on the ground, I came up with the following:

Length from top of 6 ft pole to point on ground:
L1 = sqrt (d^2 + 36)

Length from top of 8 ft pole to point on ground:
L2 = sqrt [(10-d)^d + 64] = sqrt(d^2 - 20d + 164)

Adding L1 & L2 to get the total cable length LT, I got:
LT = sqrt (d^2 - 20d + 164) + sqrt (d^2 + 36)

Taking the derivative and setting it equal to zero to find the critical numbers, I got:

LT ' = 2d^3 - 30d^2 + 400d -1640 = 0

Factoring out the 2, I got:

d^3 - 15d^2 + 200d - 820 = 0

I got stuck on trying to find the zeroes of this function. Using the rational roots test, I came up with possible rational roots at d = 2,4,5, and 10 (I ruled out negatives because d can't be negative).

I tried all these, but I did not find any of them to be a zero.

I went over my calculations a few times and I keep ending at the same dead end. If anyone can give me a clue, I'd really appreciate it.

Thanks in advance,
Paul

Setting this to zero to find critical numbers:

 

[galactus]

How did you arrive at that derivative?.

Total length is \(\displaystyle L=\sqrt{x^{2}+36}+\sqrt{(10-x)^{2}+64}\)

\(\displaystyle \frac{dL}{dx}=\frac{x-10}{\sqrt{x^{2}-10x+164}}+\frac{x}{x^{2}+36}\)

\(\displaystyle =\frac{x\sqrt{x^{2}-20x+164}+x\sqrt{x^{2}+36}-10\sqrt{x^{2}+36}}{\sqrt{(x^{2}+36)(x^{2}-20x+164)}}\)

Set to 0 and solve for x:

\(\displaystyle x\sqrt{x^{2}-20x+164}+x\sqrt{x^{2}+36}-10\sqrt{x^{2}+36}=0\)

\(\displaystyle x\sqrt{x^{2}-20x+164}=\sqrt{x^{2}+36}(10-x)\)

Square both sides:

\(\displaystyle x^{2}(x^{2}-20x+164)=(x^{2}+36)(10-x)^{2}\)

\(\displaystyle 28x^{2}+720x-3600\)

\(\displaystyle 4(x+30)(7x-30)=0\)

 

[Deleted member 4993]

Hi,

I am stuck on the following problem:

Two vertical poles of lengths 6 ft and 8 ft stand on level ground, with their bases 10 ft apart. Approximate the minimal length of cable that can reach from the top of one pole to some point on the ground between the poles, and then to the top of the other pole.

Using the result found in:

viewtopic.php?f=3&t=47272

you can show that:

\(\displaystyle \frac{6}{d} \ = \ tan(\theta) \ = \ \frac{8}{10-d}\)

d = 30/7

\(\displaystyle L = \ \sqrt{6^2+\left(\frac{30}{7}\right )^2} \ + \ \sqrt{8^2+\left(\frac{40}{7}\right)^2}\)

 

[PaulKraemer]

thank you galactus and subhotosh!

you have been a great help once again.