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 10 km east of the refinery. The cost of laying pipe is$200,000/km over land to a point P on the north bank and $400,000/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.)

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

Answer Link

abhishekprakash2625 abhishekprakash2625 · Answer 2 · 2021-10-09T21:10:21+02:00

Answer:

The point P should be [tex](10-\frac{2}{\sqrt{3} } )[/tex] or [tex]8.85\, \mathrm k\mathrm m[/tex] from the oil refinery.

Step-by-step explanation:

Let,

Cost of laying pipe over land is [tex]a[/tex].

Cost of laying pipe over river is [tex]b[/tex].

The horizontal distance between refinery and point P is [tex]x[/tex].

The horizontal distance between refinery and tank is [tex]D[/tex].

The expression for cost of the pipeline is,

[tex]C=ax+b\sqrt{(D-x)^2+4}[/tex]

Differentiate the above equation with respect to [tex]x[/tex].

[tex]\begin{aligned}C'=a+b({\frac{2(D-x)(-1)}{2\sqrt{(D-x)^2+4} })&=a-\frac{b(D-x)}{2\sqrt{(D-x)^2+4} }\\&=\frac{(a\sqrt{(D-x)^2+4})-b(D-x) }{\sqrt{(D-x)^2+4} }\\\end {aligned}[/tex]

Equate [tex]C'=0[/tex]

[tex]a\sqrt{(D-x)^2+4} -b(D-x)=0[/tex]

Now solve for [tex]x[/tex].

[tex](a^2-b^2)(D-x)^2=-4a^2\\D-x=\frac{2a}{\sqrt{b^2-a^2} } \\x=D-\frac{2a}{\sqrt{b^2-a^2} }[/tex]

Now put the values.

[tex]x=10-\frac{2\times200,000}{\sqrt{(400,000)^2-(200,000)^2} }\\x=10-\frac{2}{\sqrt{3} } \\x=8.85 \,\mathrm {km}[/tex]

Therefore, the point P should be [tex](10-\frac{2}{\sqrt{3} } )[/tex] or [tex]8.85\, \mathrm k\mathrm m[/tex] from the oil refinery.

Learn more here:

https://brainly.com/question/12965537