Differential Calculus: Maximum and Minimum Problem and Solution

An oil refinery is located on the northwestwarderly bank of a true(a) river that is 2 km wide. A cable is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying hollo is $200,000 per km over land to a academic degree P on the north bank and $400,000 per km infra the river to the tanks. To lessen the cost of the assembly line, how far from the refinery should P be located? (Round your answer to two decimal places. ) 1 year ago enunciate Ab rehearse Colorado Best execute Chosen by VotersThis is a min-max cream of tartar problem, where we want to minimize the cost answer We need a drawing of the business office 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= outdo from O to the storage tanks. Note, we could have found 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) = outperform along north shore * pipeline cost over land + withdrawnness downstairs the river * pipeline cost under land The exceed 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 graph this To find the rate of x where C(x) is minimized, we set dC/dx = 0, Reminder use the chain rule to differentiate the arcsecond term Differentiating and simplifying, we give out 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