Tuesday, January 15, 2019

Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 48

Suppose that the Oil Refinery is located 1k north of the river that is 2km wide. A pipeline is to be contructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. The cost of laying pipe is $400,000km over land to a point P on the north bank and $800,000km under the river to the tanks. Where should P be located to minimize the cost of the pipeline?



By Pythagorean Theorem, the distance between the refinery to point P is ...
d1=(6x)2+12=x212x+37

And, the distance between point P to the storage is...
d2=x2+4

Therefore, the total cost would be...
cost = 400,000x212x+37+800,000x2+4

If we take the derivative of cost, we have...
c(x)=400,000(2x122x212x+37)+800,000(2x2x2+4)

when c(x)=0

0=400,000[\cancel2(x6)\cancel2x212x+37+2(\cancel2x\cancel2x2+4)]0=x6x212x+37+2xx2+4


Solving for x, we have...
x=1.12 km

If we evaluate the cost at x=0, x=6km and x=1.12km, then...

c(0)=400,000(0212(0)+37)+800,000(02+4)c(0)=$4,033,105.012c(6)=400,00(6212(6)+37)+800,000(62+4)c(0)=$5,459,644.256c(1.12)=400,000((1.12)212(1.12)+37)+800,000((1.12)2+4)c(1.12)=$3,826,360.414

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