My diagram:
A --- storage tank
B --- point on road, 600 ft down from A
Q ---- point we want to go to, AB perpendicular to BQ
P ---- point between A and B where we want to leave the road.
Let PB = x
so we have:
PB = x, AP = 600-x, BQ = 300
L(ength) of pipeline = AP + PQ = (600-x) + √(x^2 + 300^2)
cost = 8(600-x) + 10(x^2 + 90000)^(1/2)
d cost/dx = -8 + 5(2x)(x^2 + 90000)^(-1/2) = 0 for a min of cost
10x / √(x^2 + 90000) = 8
5x = 4√(x^2 + 90000)
square both sides:
25x^2 = 16x^2 + 360000
9x^2 = 360000
x^2 = 40000
x = 200
A gas pipeline is to be constructed from a storage tank, which is right on a road, to a house which is 600 feet down the road and 300 feet back from the road. Pipe laid along the road costs $8.00 per foot, while pipe laid off the road costs $10.00 per foot. What is the minimum cost for which this pipline can be built?
i don't get how to get x =200 when i set to 0?
1 answer