Nathan Rollins
Math 1210 – Calculus
Math 1210 Pipeline Project
Dear CEO,
I am writing regarding the Vernal pipeline and its projected costs. The lowest estimated cost
would result from building pipeline through both private and BLM property. Running pipeline in the
most efficient way possible would yield a total cost of approximately $8,000,000.00. Traveling around
private land by building east through the mountain then south to the refinery would cost an additional
$100,000.00. Avoiding private property by building west, south, then east on flat land would yield an
identical result. Building in a straight line from the well to the refinery would cost more yet, totaling over
$10,000,000.00.
a) Determine the cost of running the pipeline strictly on BLM ground with two different cases, one
heading east through the mountain and then south to the refinery and the other running west, south and
then east to the refinery.
i.
Total cost for heading east through the mountain, and then south through BLM territory:
( distanceEast + distanceSouth ) * blmCostPerMile + mountainDrillingCost + studyCost
( 20 + 5 ) * 300,000 + 500,000 + 100,000 = $8,100,000.00
ii.
Total cost for heading west, south, and east through BLM territory
( distanceWest + distanceSouth + distanceEast ) * blmCostPerMile
( 1 + 5 + 21 ) * 300,000 = $8,100,000.00
b) Determine the cost of running the pipeline the shortest distance (straight line joining well to refinery
across the private ground).
i.
totalCost = shortestDistance * ( blmCostPerMile + privateGroundCostPerMile )
shortestDistance (by Pythagorean theorem) = sqrt( 20^2 + 5^2 ) = sqrt( 425 )
totalCost = sqrt( 425 ) * 500,000 = $10,307,764.07
c) Determine the optimal place to run the pipeline to minimize cost.
totalCost = privateCost + blmCost
privateCost = distanceThroughPrivateLand ( blmCostPerMile + privateGroundCostPerMile )
distanceThroughPrivateLand = sqrt( 5^2 + x^2 )
therefore, privateCost = sqrt( 25 + x^2 ) * 500,000 where x is horizontal ( west -> east ) distance
traveled through private land.
blmCost = distanceThroughBlmLand * blmCostPerMile
distanceThroughBlmLand = ( 20 – x )
therefore, blmCost = ( 20 – x ) * 300000 where x is horizontal ( west -> east ) distance traveled through
private land.
totalCost = privateCost + blmCost
therefore, totalCost = sqrt( 25 + x^2 ) * 500,000 + ( 20 – x ) * 300,000
The critical point of the cost function’s derivative will yield a possible minimum value.
(dy/dx) totalCost = 500,000( ½ ( 25+x^2 )^(-½) )
= (500,000x)/(sqrt(25+x^2)) – 300,000
0 = (500,000x)/(sqrt(25+x^2)) – 300,000
300,000 = (500,000x)/(sqrt(25+x^2))
300,000 * sqrt(25 + x^2) = 500,000x
sqrt(25 + x^2) = (5/3)x
25 + x^2 = (25/9)x^2
25 = (25/9)x^2 – x^2
25 = (16/9)x^2
x = 3.75
Minimum Cost = sqrt( 25 + 3.75^2 ) * 500,000 + ( 20 – 3.75 ) * 300,000 = $8,000,000
Minimum cost is achieved by running pipe for 6.25 miles southeast through the private land, and
then 16.25 miles east to the refinery.
6.25 mi.
16.25 mi.
d) Include a sketch of the cost function for this pipeline for configurations involving crossing some
private ground in a straight line that intersects the BLM ground to the south and then running east to the
refinery.
This graph of the cost function supports its validity, as the cost for running pipeline in a straight
line matches the cost value calculated in step B. (x=20)
Calculus, unlike most maths I have encountered up to this point in my educational career, focuses
almost entirely on maximizing efficiency. This is the material that determines whether businesses
succeed or fail, and allows for very solid prediction as well as general beneficial improvement. Whether I
use optimization to maximize workplace efficiency, or a derivative to aid in productive planning, this is
what the business world is built on, and I expect to find many applications for this material in the future.