MHB Minimize Total Length of Cables: Estimate Minimum Value?

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Cable Length
AI Thread Summary
To minimize the total length of cables linking point P to points A, B, and C, the function L(x) is defined using the Pythagorean theorem. The derivative L'(x) is calculated and set to zero to find critical values, leading to the application of Newton's method for approximation. Starting with an initial guess, the iterations converge to a value of approximately 1.407 for x. The second derivative test confirms this is a global minimum, resulting in a minimum cable length of approximately 9.35 meters. The calculations demonstrate an effective method for estimating minimum values in optimization problems.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

How to estimate the minimum value?

A point P needs to be located somewhere on the line AD so that the total length L of cables linking P to the points A, B, and C is minimized (see the figure). Estimate the minimum value of L to two decimals.

View attachment 2123

So far I have:

dL/dx=1 + ((x-5)/(sqrt(29-10 x+x^2))) + ((x-5)/(sqrt(50-10 x+x^2)))

What is the next step??

I have posted a link there to this thread so the OP can view my work.
 

Attachments

  • lavenderv.png
    lavenderv.png
    3.1 KB · Views: 119
Mathematics news on Phys.org
Hello Lavender V,

Let's let $0\le x\le 5$ represent the length of segment $\overline{DP}$ in meters. And so the total length of cable as a function of $x$ is then obtained via the Pythagorean theorem as:

$$L(x)=(5-x)+\sqrt{x^2+2^2}+\sqrt{x^2+3^2}$$

Now, we may differentiate with respect to $x$ and equate the result to zero to obtain the critical value(s):

$$L'(x)=-1+\frac{x}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+9}}=0$$

Now, since we are thankfully allowed to approximate, we may use Newton's method here to approximate the root of this equation.

Let:

$$f(x)=-1+\frac{x}{\sqrt{x^2+4}}+\frac{x}{\sqrt{x^2+9}}$$

We need to compute the first derivative, and so for the last two terms, consider:

$$g(x)=\frac{x}{\sqrt{x^2+k}}$$

Using the quotient, power and chain rules, we obtain:

$$g'(x)=\frac{\sqrt{x^2+k}(1)-x\dfrac{x}{\sqrt{x^2+k}}}{\left(\sqrt{x^2+k} \right)^2}=\frac{k}{\left(x^2+k \right)^{\frac{3}{2}}}$$

And so we may conclude that:

$$f'(x)=\frac{4}{\left(x^2+4 \right)^{\frac{3}{2}}}+\frac{9}{\left(x^2+9 \right)^{\frac{3}{2}}}$$

Newton's method gives us the recursion:

$$x_{n+1}=x_{n}-\frac{f\left(x_n \right)}{f'\left(x_n \right)}$$

And so using the definition of $f$ and its derivative, we have:

$$x_{n+1}=x_{n}-\frac{-1+\dfrac{x_n}{\sqrt{x_n^2+4}}+\dfrac{x_n}{\sqrt{x_n^2+9}}}{\dfrac{4}{\left(x_n^2+4 \right)^{\frac{3}{2}}}+\dfrac{9}{\left(x_n^2+9 \right)^{\frac{3}{2}}}}$$

Now, using a computer or CAS, and $x_0=1$, we find:

$$x_1\approx1.36825580018177$$

$$x_2\approx1.40658793851904$$

$$x_3\approx1.40699724822532$$

$$x_4\approx1.40699729457262$$

$$x_5\approx1.40699729457262$$

This is accurate to 15 digits, more than enough to get an approximation for the minimum of $L$ to two decimal places. Now, if we note that:

$$f'(x)=L''(x)>0$$

For all real values of $x$, then by the second derivative test, we may conclude that this critical value is at the global minimum, and we need not check the boundaries.

Thus, we may conclude:

$$L_{\min}\approx L(1.40699729457262)\approx9.35\text{ m}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top