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 3 km east of the refinery. The cost of laying pipe is $400,000 per km over land to a point P on the north bank and $800,000 per km under the river to the tanks. To minimize the cost of the pipeline, how far from the refinery should P be located? (Round your answer to two decimal places.)

Let's d be the distance in km from the oil refinery to point P. So the horizontal distance from P to the storage is 3 - d and the vertical distance is 2. Hence the diagonal distance is:

[tex]\sqrt{(3 - d)^2 + 2^2} = \sqrt{(3 - d)^2 + 4}[/tex]

So the cost of laying pipe under water with this distance is

[tex]800000\sqrt{(3 - d)^2 + 4}[/tex]

And the cost of laying pipe over land from the refinery to point P is 400000d. Hence the total cost:

[tex]800000\sqrt{(3 - d)^2 + 4} + 400000d[/tex]

We can find the minimum value of this by taking the 1st derivative and set it to 0

[tex]800000\frac{2*0.5*(3-d)(-1)}{\sqrt{(3 - d)^2 + 4}} + 400000 = 0[/tex]

We can move the first term over to the right hand side and divide both sides by 400000

[tex]1 = 2\frac{3 - d}{\sqrt{(3 - d)^2 + 4}}[/tex]

[tex]\sqrt{(3 - d)^2 + 4} = 6 - 2d[/tex]

From here we can square up both sides

[tex](3 - d)^2 + 4 = (6 - 2d)^2[/tex]

[tex]9 - 6d + d^2 + 4 = 36 - 24d + 4d^2[/tex]

[tex]3d^2-18d+27 = 0[/tex]

[tex]d^2 - 6d + 9 = 0[/tex]

[tex](d - 3)^2 = 0[/tex]

[tex]d -3 = 0[/tex]

d = 3

So the cost of pipeline is minimum when P is exactly 3km east from the oil refinery.