the actual costs are not important; just their ratio ... water:land = 2:1
If P is located x km east, then the cost
c = 2√(1+x^2) + 1(7-x)
dc/dx = 2x/√(1+x^2) - 1
dc/dx=0 at x = 1/√3
An oil refinery is located on the north bank of a straight river that is 1 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 7 km east of the refinery. The cost of laying pipe is $500,000/km over land to a point P on the north bank and $1,000,000/km under the river to the tanks. To minimize the cost of the pipeline, how far (in km) from the refinery should P be located?
3 answers
s is length of underwater pipe, hypotenuse
c = cost in units of $100,000 / km
c = 5 x + 10 s
dc/dx = 5 + 10 ds/dx
zero when ds/dx = -0.5
geometry:
1^2 +(7-x)^2 = s^2
1 + 49 - 14 x + x^2 = s^2
s^2 = x^2 - 14 x + 50
2 s ds/dx = 2 x - 14
2 s (-0.5) = 2 x - 14
2 x + s = 14
s = 14 - 2x
1^2 + (7-x)^2 = s^2 = (14-2x)^2
1 + 49 - 14 x^2 = 196 - 56 x + 4 x^2
18 x^2 - 105 x + 146 = 0
I get x = 3.55 or x = 2.29
may have an arithmetic error, working fast
c = cost in units of $100,000 / km
c = 5 x + 10 s
dc/dx = 5 + 10 ds/dx
zero when ds/dx = -0.5
geometry:
1^2 +(7-x)^2 = s^2
1 + 49 - 14 x + x^2 = s^2
s^2 = x^2 - 14 x + 50
2 s ds/dx = 2 x - 14
2 s (-0.5) = 2 x - 14
2 x + s = 14
s = 14 - 2x
1^2 + (7-x)^2 = s^2 = (14-2x)^2
1 + 49 - 14 x^2 = 196 - 56 x + 4 x^2
18 x^2 - 105 x + 146 = 0
I get x = 3.55 or x = 2.29
may have an arithmetic error, working fast
Use oobleck solution, variable choices much smarter.
1 answer