Question

An offshore oil well is located in the ocean at a point 𝑾, which is 5 𝑚𝑖𝑙𝑒𝑠 from the closest shore-point 𝑨 on a straight shoreline. The oil is to be piped to a shore-point 𝑩 that is 8 𝑚𝑖𝑙𝑒𝑠 from 𝑨 by piping it on a straight line underwater from 𝑾 to some shore-point 𝑷 between 𝑨 and 𝑩 and then on to 𝑩 via a pipe along the shoreline. If the cost of laying pipe is $100,000.00 per mile under water and $75,000.00 per mile over land, where should point 𝑷 be located to minimize the cost of laying the pipe?

Answer 1

Cost of pipe = Cost under water + Cost on land

$=100000x+75000y$

Using Pythagoras theorem, we get:

$(5)^2+(8-y)^2=x^2$

We need to change this equation in terms of one variable:

$=100000(\sqrt{(5)^2+(8-y)^2})+75000y\\$

We need to find the cost price and then minimize it.

Let $C(y)=100000(\sqrt{(5)^2+(8-y)^2})+75000y$

$\therefore C'(y)=100000\times \frac{1}{(\sqrt{(5)^2+(8-y)^2})}\times (-2(8-y))+75000$

Put $C'(y)=0$ , we get:

$\Rightarrow100000\times \frac{1}{(\sqrt{(5)^2+(8-y)^2})}\times (-2(8-y))+75000=0$

$\Rightarrow \frac{-200(8-y)}{(\sqrt{(5)^2+(8-y)^2})}+75=0\\ \Rightarrow -200(8-y)=-75{(\sqrt{(5)^2+(8-y)^2})}\\ \Rightarrow 8(8-y)=3{(\sqrt{(5)^2+(8-y)^2})}\\$

Squaring both sides, we get:

$64(8-y)^2=9(25+(8-y)^2)\\ \Rightarrow 55(8-y)^2=225\\ \Rightarrow (8-y)^2=\frac{45}{11}\\ \Rightarrow y=5.97$

So, $8-y=8-5.97=2.03$ miles right of point A.