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September 2, 2014

September 2, 2014

Posted by **Aaron** on Monday, November 30, 2009 at 10:28pm.

- Calc -
**MathMate**, Tuesday, December 1, 2009 at 12:35amLet x be the distance between P and a point on the north bank directly opposite to the tanks.

The length of the pipe under water,

L1(x) = √(2²+x²).

The length of the pipe on land,

L2(x) = 3-x

Cost of laying the pipes,

C(x) = 600000 L1(x) + 300000 L2(x)

Differentiate C(x) with respect to x and equate the derivative to zero.

Solve for x.

The distance from the refinery to P is (3-x).

I get about 1.8 km.

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