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A small resort is situated on an island off a part of the coast of Mexico that has a perfectly straight north-south shoreline. The point P on the shoreline that is closest to the island is exactly 6 miles from the island. Ten miles south of P is the closest source of fresh water to the island. A pipeline is to be built from the island to the source of fresh water by laying pipe underwater in a straight line from the island to a point Q on the shoreline between P and the water source, and then laying pipe on land along the shoreline from Q to the source. It costs 1.6 times as much money to lay pipe in the water as it does on land. How far south of P should Q be located in order to minimize the total construction costs?

Hint: You can do this problem by assuming that it costs one dollar per mile to lay pipe on land, and 1.6 dollars per mile to lay pipe in the water. You then need to minimize the cost over the interval [0,10] of the possible distances from P to Q.

The money derivatives over land is $1 per mile, and the money derivative over water is $1.6 per mile. x has to be between 0 and 10, the maximum possible distance underwater is sqrt(136) miles. However, that will have a rather high cost. I know that the distance will be closer to 0 than to 10. The equation will most likely be a positive quadratic equation. How will I find this equation, the distance, and the cost at this location.

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  • $\begingroup$ Can you write an equation for the total distance travelled, or better still - for the cost, given any value of $x$? Also are you familiar with calculus? $\endgroup$ – Macavity Apr 21 '13 at 3:31
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We calculate the cost $C(x)$ of going underwater to a point $x$ miles south of $P$, and then heading on land to the water source.

Draw a picture. By the Pythagorean Theorem, the straight line distance from the island to a point $x$ miles South of $P$ is $\sqrt{6^2+x^2}$.

Then the distance along the shore to the water source is $10-x$, Taking costs into account, we find that $$C(x)=1.6\sqrt{6^2+x^2}+ 10-x.$$ We want to minimize $C(x)$, given that $0\le x\le 10$. As usual, we find where $C'(x)=0$ between $x=0$ and $x=10$. Then we find out which is best, the point(s) that give $C'(x)=0$, and the endpoints $x=0$ and $x=10$.

If you are comfortable with trigonometric functions, it may be easier to let the island be the point $A$, and the point along the shore that we head for be $B$. Let $\theta=\angle PAB$. Then the distance under water from $A$ to $B$ is $6\sec\theta$, and the distance along the shore from $B$ to the water source is $10-6\tan\theta$. So our cost in terms of $\theta$ is $f(\theta)$, where $$f(\theta)=9.6\sec\theta +(10-\tan\theta).$$ Minimize $f(\theta)$.

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  • $\begingroup$ @Macavity: Thanks! Fixed. $\endgroup$ – André Nicolas Apr 21 '13 at 4:58
  • $\begingroup$ following your approach, should be: $$f(\theta)=9.6\sec\theta +(10-6\tan\theta).$$ $\endgroup$ – Macavity Apr 21 '13 at 7:40
  • $\begingroup$ @Macavity: Thanks again. $\endgroup$ – André Nicolas Apr 21 '13 at 14:19

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