# what is the cheapest way to lay the pipe?

• October 29th 2012, 01:01 AM
what is the cheapest way to lay the pipe?
Hi guys!

I attached a digram explaining the equation and I'd really like to know how to get the cheapest rout :D
This question has been pondering in my mind for a long lime and I've never known how to work it out
what is the cheapest way of laying the pipe?
also what is the distance between B & X to get the cheapest price?

Edit: Please keep the distance B to X rounded to 3 decimals as that is the closest meter :)
• October 29th 2012, 01:05 AM
Re: what is the cheapest way to lay the pipe?
Just by working it out by trial and error got me BX=5.8km and price was $3,165,400. • October 29th 2012, 01:26 AM MarkFL Re: what is the cheapest way to lay the pipe? The distance in km of the underwater portion is: $\sqrt{x^2+10^2}$ The distance in km of the underground portion is: $16-x$ And so, the total cost in thousands of dollars is: $C(x)=190\sqrt{x^2+10^2}+95(16-x)$ where $0\le x\le16$ To find the minimum, we need to differentiate this cost function with respect to $x$ and equate to zero: $C'(x)=\frac{190x}{\sqrt{x^2+10^2}}-95=\frac{95(2x-\sqrt{x^2+10^2})}{\sqrt{x^2+10^2}}=0$ $2x=\sqrt{x^2+10^2}$ $4x^2=x^2+10^2$ $3x^2=10^2$ $x^2=\frac{10^2}{3}$ $x=\frac{10}{\sqrt{3}}$ The first derivative test shows we have a minimum at this point, so the minimum cost is: $C\left(\frac{10}{\sqrt{3}} \right)=950\sqrt{3}+1520\approx3165.44827$ Thus, the minimum cost is about$3,165,448.27.
• October 29th 2012, 01:36 AM
MarkFL
Re: what is the cheapest way to lay the pipe?
Since you posted in the algebra forum, I assume this is not homework, but you are just curious how it is to be solved. You are very close with your trial and error estimate, by the way, although your computed cost is wrong, but your value for x is very close.

This is a problem typically given to Calc I students.
• October 29th 2012, 01:50 AM