# Minimum problem...with distances.

• Jan 4th 2009, 04:31 PM
Jukixel
Minimum problem...with distances.
This one requires some drawing...I think. A diagram maybe? ^^"

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)
• Jan 4th 2009, 05:14 PM
Soroban
Hello, Jukixel!

Quote:

Town A is 11 miles from a straight river and town B is 6 miles from the same river.
The distance from town A to town B is 13 miles.
A pumping station is to be built along the river to supply water to both towns.
Where should the pumping station be built so that the sum of the distances
from the pumping station to the towns is a minimum?

There are two stages to this problem . . .

First, find the "horizontal" distance between the two towns.

In the diagram below, right triangle $ABE$ has side 5 and hypotenuse 13.
. . Hence: . $AB \,=\,CD \,=\,12$
Code:

    A * - - - - - - - - - - - - - * B       |                          :       |        *                : 5       |            13    *      :   11 |                          * E       |                          |       |                          |       |                          | 6       |                          |     - * - - - - - - - - - - - - - * -       C                          D

Then the diagram looks like this . . .
Code:

    A *       | *       |  *       |    *   11 |      *                  * E       |        *              * |       |          *          *  |       |            *      *    | 6       |              *  *      |       * - - - - - - - - *- - - - *       C        x        P  12-x  D

The pumping station is at $P.$

Let $CP \,=\,x \quad\Rightarrow\quad PD \,=\,12-x$

In right triangle $ACP\!:\;\;AP \,=\,\sqrt{x^2+11^2}$

In right triangle $EDP\!:\;\;PE \,=\,\sqrt{(12-x)^2+6^2}$

The total distance from $P$ to the towns is:

. . $D \:=\:\sqrt{x^2+121} + \sqrt{x^2 - 24x + 180}$

And that is function you must minimize . . .

• Jan 4th 2009, 05:15 PM
mr fantastic
Quote:

Originally Posted by Jukixel
This one requires some drawing...I think. A diagram maybe? ^^" Mr F says: You think??

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)

A diagram is essential. Draw one.

Now find the horizontal distance between each town. From you diagram you should see that the line segment joining A and B is the hypotenuse of a right-triangle. Use Pythagoras' theorem to get that the horizontal distance between A and B is 12 miles (alternatively, you are familiar with your Pythagorean triples and will have no trouble recognising the 5-12-13 right-triangle).

Let the pumping station P be x miles along the river from the perpendicular line joing A to the river.

Then $d\overline{AP}^2 = 11^2 + x^2$ and $d\overline{BP}^2 = 6^2 + (12 - x)^2$.

Therefore $L = d\overline{AP} + d\overline{BP} = \, ....$

Edit: Is this really a pre-calculus question, that is, you're expected to use technology to find the value of x that minimises the distance ....?

Otherwise it requires calculus .....
• Jan 5th 2009, 10:18 AM
HallsofIvy
Quote:

Originally Posted by Jukixel
This one requires some drawing...I think. A diagram maybe? ^^"

Town A is 11 miles from a straight river and town B is 6 miles from the same river. The distance from town A to town B is 13 miles. A pumping station is to be built along the river to supply water to both towns. Where should the pumping station be built so that the sum of the distances from the pumping station to the town is a minimum?

Thanks for the help. =)

I see that both Soroban and Mr. Fantastic are assuming the two towns are on the same side of the river. With that, it follows that distance "along the river" from A to B (from the point where the perpendicular from A meets the river to the point where the perpendicular is 12 miles.

But I would do this in a completely different way (especially if you are cannot use "Calculus"): using a "reflection". That is, imagine that B' is 6 miles from the river but on the opposite side of the river from B. Obviously (I hope it is obvious!) the shortest distance from A to B' is a straight line. We can determine where that line crosses the river by "similar triangles"- the triangle formed by A, foot of perpendicular from A to the river, point where the straight line from A to B' crosses the river is similar to the corresponding B' triangle. That follows from the "vertical angles" theorem- where lines cross (the line from A to B' and the river) angles across from one another are congruent.

If we let the distance from where the perpendicular from A hits the river to the point where the line AB' crosses the river be "x" then the distance from that point to where the perpendicular from B hits the river is 12- x. So we have $\frac{11}{x}= \frac{6}{12-x}$ so 11(12- x)= 132- 11x= 6x or 5x= 132 and x= 132/5= 26.4 miles.

Now you can show that this point is also correct for the shortest total distance between A and B. The two triangles, with B' and B are congruent so the distance from that crossing point on the river to B' is the same as from that point to B.
• Jan 5th 2009, 11:39 AM
mr fantastic
Quote:

Originally Posted by HallsofIvy
I see that both Soroban and Mr. Fantastic are assuming the two towns are on the same side of the river. With that, it follows that distance "along the river" from A to B (from the point where the perpendicular from A meets the river to the point where the perpendicular is 12 miles.

But I would do this in a completely different way (especially if you are cannot use "Calculus"): using a "reflection". That is, imagine that B' is 6 miles from the river but on the opposite side of the river from B. Obviously (I hope it is obvious!) the shortest distance from A to B' is a straight line. We can determine where that line crosses the river by "similar triangles"- the triangle formed by A, foot of perpendicular from A to the river, point where the straight line from A to B' crosses the river is similar to the corresponding B' triangle. That follows from the "vertical angles" theorem- where lines cross (the line from A to B' and the river) angles across from one another are congruent.

If we let the distance from where the perpendicular from A hits the river to the point where the line AB' crosses the river be "x" then the distance from that point to where the perpendicular from B hits the river is 12- x. So we have $\frac{11}{x}= \frac{6}{12-x}$ so 11(12- x)= 132- 11x= 6x or 5x= 132 and x= 132/5= 26.4 miles.

Now you can show that this point is also correct for the shortest total distance between A and B. The two triangles, with B' and B are congruent so the distance from that crossing point on the river to B' is the same as from that point to B.

The old unconscious assumption trick. Second time this week I've fallen for it.