1. Optimization problem

Hello, I am stuck on a problem and need a bit of help.

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000/km over land to a point P on the north bank and$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should the point P be located?

Here is what I have so far.. (sorry for the drawing, I'm new to this and don't know of any better way to do it) EDIT: Drawing didn't transfer very well from the edit box

I know that
$Cost = 400,000x + 800,000y$

and
$y = sqrt(2^2 +(6-x)^2)$

and I know that to find the min we need to find $dC/dx$ and set equal to 0 but when I do this I get stuck when solving for x. Maybe I'm messing up my algebra, I don't know, but any help would be appreciated.

Thanks

2. Originally Posted by Zion
Hello, I am stuck on a problem and need a bit of help.

An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $400,000/km over land to a point P on the north bank and$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should the point P be located?

Here is what I have so far.. (sorry for the drawing, I'm new to this and don't know of any better way to do it) EDIT: Drawing didn't transfer very well from the edit box

I know that
$Cost = 400,000x + 800,000y$

and
$y = sqrt(2^2 +(6-x)^2)$

and I know that to find the min we need to find $dC/dx$ and set equal to 0 but when I do this I get stuck when solving for x. Maybe I'm messing up my algebra, I don't know, but any help would be appreciated.

Thanks
Hi Zion,

for simplicity let the cost be 4x+8y hundred thousand dollars.

$\frac{d}{dx}\left(8\sqrt{(6-x)^2+4}+4x\right)=8\frac{du}{dx}\ \frac{d}{du}u^{\frac{1}{2}}+4$

$=4+(8)\frac{1}{2}u^{-\frac{1}{2}}\ \frac{dv}{dx}\ \frac{d}{dv}v^2=4+4\ \frac{1}{\sqrt{(6-x)^2+4}}(-1)(2)(6-x)$

where

$u=(6-x)^2+4$

$v=6-x$

Setting this derivative to zero gives

$4+4\ \frac{1}{\sqrt{(6-x)^2+4}}2(x-6)=0$

$\frac{8(x-6)}{\sqrt{(6-x)^2+4}}=-4$

$\frac{8(6-x)}{\sqrt{(6-x)^2+4}}=4$

$2(6-x)=\sqrt{(6-x)^2+4}$

$4(6-x)^2=(6-x)^2+4$

$3(6-x)^2=4$

$(6-x)^2=\frac{4}{3}$

$6-x=\frac{2}{\sqrt{3}}$

$x=6-\frac{2}{\sqrt{3}}=\frac{6\sqrt{3}-2}{\sqrt{3}}$

Now you can calculate y from $(6-x)^2=\frac{4}{3}$

and calculate the minimum cost from these x and y.

3. Thanks so much!

I was on the right track but kept getting tripped up in my algebra