Answer:
The point P should be located at 0,51 Km from point B, to minimize costs
Step-by-step explanation:
The oil refinery ( point A ) the opposite point ( B) in the south bank, and the storage tank (point C) these three points shape a right triangle. Let´s call x the distance between point A and point P ( the arriving point of the pipeline in the south bank ) then:
Cost = cost of pipeline in-ground* ( 6 - x ) + cost of pipeline in water* L
where L is the hypothenuse of the right triangle ( A , B , P )
Cost ($) = 400000 * ( 6 - x ) + 800000 * √ 2² + x²
C (x) = 400000* ( 6 - x ) + 800000* √4 + x²
C(x) = 2400000 - 400000*x + 800000* √4 + x²
Tacking derivatives on both sides of the equation
C´(x) = - 400000 + 800000* ( 2x)/√4 + x²
C´(x) = 0 - 400000 + 1600000*x / √4 + x² = 0
- 400000* (√4 + x² ) + 1600000*x = 0
- (√4 + x² ) + 4* x = 0
- (√4 + x² ) = - 4*x
Squaring both sides
4 + x² = 16*x²
15*x² = 4
x² = 4/ 15
x = ± √0,27
x = ± 0,51
We should dismiss the negative solution
x = 0,51 Km
The point P should be located at 0,51 Km from point B