optimization: a pipeline for transporting oil...

[zebra]

A pipeline for trasporting oil will connect two points A and B that are 3 miles apart and on opposite banks of a straight river 1 mile wide. Part of the pipeline will run under water from A to a point C on the opposite bank, and then above ground from C to B. If the cost per mile of running the pipeline under water is four times the cost per mile of running it above ground, find the location of C that will minimize the cost(disregard the slope of the river bed).

i determined that the distance from C to B will equal "c" in terms of cost and the distance from A to C will equal "4c" for later use.

i looked at the picture and saw a triangle with, according to our knowledge, three unequal sides so i split it in two in order to make two right-angled trangles.

i dont know exactly where to go from here in order to find a cst function with regard to the position of C

 

[soroban]

Hello, zebra!

A pipeline for trasporting oil will connect two points A and B that are 3 miles apart
. . and on opposite banks of a straight river 1 mile wide.
Part of the pipeline will run under water from A to a point C on the opposite bank,
. . and then above ground from C to B.
If the cost per mile of underwater pipe is four times the cost of above-ground pipe,
. . find the location of C that will minimize the cost.

    A *
      |  *
      |     *     ______
    1 |        * √x² + 1
      |           *
      |              *
    --*-----------------*-------------*--
      D       x         C     3-x     B
      : - - - - - - - 3 - - - - - - - :

The nearest point on the shore is \(\displaystyle D.\)
. . Then: \(\displaystyle \,AD\,=\,1,\;DB\,=\,3.\)

Let \(\displaystyle x\,=\,DC\), then \(\displaystyle 3-x\,=\,CB\)


There will be \(\displaystyle 3\,-\,x\) miles of above-ground pipe.
. . At \(\displaystyle c\) dollars per mile, the cost is: \(\displaystyle \,c(3\,-\,x)\) dollars.

From Pythagorus, we have: \(\displaystyle \sqrt{x^2\,+\,1}\) miles of underwater pipe.
. . At \(\displaystyle 4c\) dollars per mile, the cost is: \(\displaystyle \,4x\sqrt{x^2\,+\,1}\) dollars.


Hence, the total cost is: \(\displaystyle \:C\:=\:c(3\,-\,x)\,+\,4c\left(x^2\,+\,1\right)^{\frac{1}{2}}\)

And that is the function we must minimize . . .

 

[zebra]

wow thx but when u minimize that there are 2 variables...but then again "c" is just a constant right...so would the answer be in terms of "c"? heres what iv done so far

C= c(3-x) + 4c(x^2 +1)^(1/2)

dC/dx= -c + (2c)(2x)(x^2 +1)^(-1/2)

dC/dx= 4cx - c = 0
(x^2 +1)^(1/2)


am i going in the right direaction or did i just make a mess?

 

[soroban]

Hello again, zebra!

When u minimize that there are 2 variables
. . but then again "c" is just a constant, right? . . . Right!
so would the answer be in terms of "c"? . . no, the "c" drops out.

Here's what i've done so far

\(\displaystyle C\;= \;c(3\,-\,x)\,+\,4c(x^2\,+\,1)^{\frac{1}{2}}\)

\(\displaystyle \frac{dC}{dx}\;=\;-c\,+\,(2c)(2x)(x^2\,+\,1)^{-\frac{1}{2}}\)

\(\displaystyle \frac{dC}{dx} \;= \;\frac{4cx}{(x^2\,+\,1)^{\frac{1}{2}}}\,-\,c \;=\;0\)

Am i going in the right direction? . . . You're doing great!

We have: \(\displaystyle \,\frac{4cx}{\sqrt{x^2\,+\,1}} \;=\;c\;\;\Rightarrow\;\; 4cx\;=\;c\sqrt{x^2\,+\,1}\)

Square both sides: \(\displaystyle \:16c^2x^2\;=\;c^2(x^2\,+\,1)\;\;\Rightarrow\;\;15c^2x^2\:=\:c^2\;\;\Rightarrow\;\;x^2\:=\:\frac{1}{15}\)

Therefore: \(\displaystyle \,x\:=\:\frac{1}{\sqrt{15}} \:\approx\:0.2582\) miles.

 

[zebra]

yay!!! thanx so much