Differential Calculus: Maximum and Minimum Problem and Solution

An petroleum refinery is located on the nitrogen beach of a straight river that is 2 km wide. A cable is to be constructed from the refinery to reposition tanks located on the south imprecate of the river 6 km east of the refinery. The constitute of laying pipe is $200,000 per km oer land to a point P on the northerly bank and $400,000 per km under the river to the tanks. To pick at the hail of the pipeline, how distant from the refinery should P be located? (Round your final result to two decimal places. ) 1 social class ago Report Abuse atomic number 27 Best Answer Chosen by VotersThis is a min-max calculus problem, where we want to minimize the cost solve We need a drawing of the situation see https//docs. google. com/drawings/d/1PvkU where R is the refinery, O will be the x-axis origin, P is the point on the north bank, and x= duration from O to the retentiveness tanks. Note, we could have put R at the origin, but the algebra is a little simpler this way The cost C(x) of the pipeline as a function of x is C(x) = outstrip along north shore * pipeline cost all over land + distance under the river * pipeline cost under land The distance along the north shore is 6-xThe distance (by Pythagorean theorem) under the water is sqrt( 22 + x2) So, C(x) = (6-x)*200000 + sqrt(4 + x2) * 400000 You should interpret this To find the value of x where C(x) is minimized, we commemorate dC/dx = 0, Reminder use the chain tower to differentiate the second term Differentiating and simplifying, we get dC/dx = C(x) = -200000 + 400000x/ sqrt(4+x2) = 0 400000x / sqrt(4+x2) = 200000 400000x/200000 = sqrt(4+x2) Squaring both sides, we get 4x2= 4 + x2 x = sqrt(4/3) = 1. 15 So the distance from the refinery to point P is 6-x = 4. 85 km