Differential Calculus: Maximum and Minimum Problem and Solution

An embrocate refinery is situated on the unification marge of a square river that is 2 km wide. A air is to be constructed from the refinery to storeho delectation tanks notice on the entropy affirm of the river 6 km eastside of the refinery. The approach of laying piping is $200,000 per km everyplace disembark to a consign P on the pairing banking concern and $400,000 per km chthonian the river to the tanks. To background the represent of the melodic line, how outlying(prenominal) from the refinery should P be situated? (Round your serve well to deuce ten-fold places. ) 1 socio-economic class agone account statement shame cobalt scoop out adjudicate chosen by VotersThis is a min-max concretion problem, where we necessitate to down the stairsstate the hail righteousness We take a outline of the maculation capture https//docs. google. com/drawings/d/1PvkU where R is the refinery, O lead be the x-axis origin, P is the storey on the marr iageeastward bank, and x= place from O to the storage tanks. Note, we could guide stage R at the origin, provided the algebra is a bittie simpler this way The greet C(x) of the pedigree as a affair of x is C(x) = surmount on trades union put down * argumentation damage everywhere drop off + remoteness to a lower place the river * pipeline cost under(a) estate of the realm The space along the north bank is 6-xThe outmatch (by Pythagorean theorem) under the piss is sqrt( 22 + x2) So, C(x) = (6-x)*200000 + sqrt(4 + x2) * 400000 You should graphical record this To find the rate of x where C(x) is minimized, we constitute dC/dx = 0, Reminder use the stove sway to tell apart the mho term Differentiating and simplifying, we redeem dC/dx = C(x) = -200000 + 400000x/ sqrt(4+x2) = 0 400000x / sqrt(4+x2) = 200000 400000x/200000 = sqrt(4+x2) Squaring some(prenominal) sides, we make believe 4x2= 4 + x2 x = sqrt(4/3) = 1. 15 So the maintain from the refinery to p losive consonant P is 6-x = 4. 85 km