Question

1. -/1 points details my notes ask your teacher practice another an oil refinery is located on the north bank of a straight river that is 2 km wide. a pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 8 km east of the refinery. the cost of laying pipe is $400,000/km over land to a point p on the north bank and $800,000/km under the river to the tanks. to minimize the cost of the pipeline, how far (in km) from the refinery should p be located? (round your answer to two decimal places.) km

Explanation:

Step1: Set up the cost function

Let \(x\) be the distance (in km) from the point directly across the river on the north bank to point \(P\). The length of the pipeline over - land is \(L_{over - land}=\sqrt{x^{2}+4}\) (using the Pythagorean theorem for the right - triangle with height \(2\) km (width of the river) and base \(x\) km). The length of the pipeline under - land is \(L_{under - land}=(8 - x)\) km.

The cost function \(C(x)\) is given by:

$$C(x)=800000\sqrt{x^{2}+4}+400000(8 - x)$$

$$C(x)=800000(x^{2}+4)^{\frac{1}{2}}+3200000-400000x$$

Step2: Find the derivative of the cost function

Using the chain rule, if \(y=(x^{2}+4)^{\frac{1}{2}}\), then \(y^\prime=\frac{2x}{2\sqrt{x^{2}+4}}=\frac{x}{\sqrt{x^{2}+4}}\)

The derivative of \(C(x)\) is:

$$C^\prime(x)=800000\times\frac{x}{\sqrt{x^{2}+4}}-400000$$

$$C^\prime(x)=\frac{800000x}{\sqrt{x^{2}+4}}-400000$$

Step3: Set the derivative equal to zero and solve for \(x\)

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Answer:

\(1.15\) km