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August 1, 2014

August 1, 2014

Posted by **ibranian** on Saturday, August 11, 2012 at 11:13pm.

- math -
**Steve**, Sunday, August 12, 2012 at 2:12pmDraw a diagram. Let P be the point on shore where the pipe goes underwater.

Then if D is the drill rig, and R is the refinery,

x^2 = 12^2 + (20-y)^2

and the cost is

c = 50x+30y

c = 50√(144 + (20-y)^2) + 30y

dc/dy = 50(y-20)/√(y^2 - 40y + 544) + 30

dc/dy=0 when

y = 11

so, x = 15

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