Finding lowest value measures between three locations

Question

I've made an image of what the geography of my problem looks like:

Essentially, there are two "chemical plants" (A and B) located on a road 12 miles apart. The pollution from plant A is given by the following equation:

$\dfrac{K}{x^2 + 10}$ (for some constant K)

The pollution from plant B (at x miles from plant B) is 1/4 that of A.

There's also a third plant, C, which is located on the perpendicular road (the one branching off of Road A-B on the map). Plant C is 5 miles from A and 10 miles from B. The pollution from plant C is twice that of B.

Basically, what I am trying to find is the point on Road A-B where the pollution count from the three plants is minimal. I'm not really sure where to even begin with this.

Edit: Trying to solve this problem...

The pollution from Plant B must be $\dfrac{K}{4x^2 + 40}$, and from Plant C must be $\dfrac{K}{2x^2 + 20}$. Adding the three equations would yield the following:

$\dfrac{K}{x^2 + 10}$ + $\dfrac{K}{4x^2 + 40}$ + $\dfrac{K}{2x^2 + 20}$ = $\dfrac{K + 4K + 2K}{x^2 + 10}$ = $\dfrac{7K}{x^2 + 10}$

Finding the derivative of this would be? Not sure I'm doing this right.

$(\dfrac{7K}{x^2 + 10})'$ = $-\dfrac{14K}{(x^2 + 10)^2}$

Answer 1

First, let's find the coordinates of Plant $C$. Let $h$ be the height of the triangle formed by $A$, $B$, and $C$. Let $e$ be the distance from $A$ to the road junction of road $AB$ and the perpendicular road.

Then, by the Pythagorean Theorem $$\eqalign{ h^2&=25 -e^2\cr h^2&=100-(12-e)^2 } $$ Solving the above system, we obtain $e={23\over 8}$ and $h=\sqrt{25-(23/8)^2} =\sqrt{1071/64}$.

So, with the origin at plant $A$, the coordinates of $C$ are $(e,-h )$.

Now let $a$ be a point on the road from $A$ to $B$ that is $x$ miles from plant $A$.

Then

$\ \ \ \ \ $the distance from $a$ to $B$ is $12-x$

and

$\ \ \ \ \ $the distance from $a$ to $C$ is $ \sqrt{ (x-e)^2+ h^2 } =\sqrt{(x-(23/8))^2+ {1071/64}}$.

The pollution from plant $A$ is $$ P_A(x) = {K\over x^2+10}. $$

The pollution from plant $B$ is one fourth the pollution from plant $A$ at a point $12-x$ miles from plant $A$ $$ P_B(x)={1\over4}{K\over (x-12)^2+10 }. $$

The pollution from plant $C$ is twice the pollution from plant $B$ at a point $\sqrt{(x-(23/8))^2+ {1071/64}}$ miles from plant $B$. But the pollution from plant $B$ at a point $\sqrt{(x-(23/8))^2+ {1071/64}}$ miles from plant $B$ is one forth the pollution from plant $A$ at a point $\sqrt{(x-(23/8))^2+ {1071/64}}$ miles from plant $A$. So $$ P_C(x) =2\cdot{1\over 4} { K\over (\sqrt{(x-(23/8))^2+ {1071/64}})^2+10 } = {1\over 2} { K\over ( {(x-(23/8))^2+ {1711/64}} }. $$

The total pollution at $a$ is $P(x)=P_A(x)+P_B(x)+P_C(x)$: $$ P(x)= {K\over x^2+10}+{1\over4}{K\over (x-12)^2+10 }+ {1\over 2} { K\over ( {(x-(23/8))^2+ {1711/64}} } $$

You need to minimize $P$ over the interval $[0,12]$.

I would surmise that this would be very difficult to do by hand. Should you be so inclined to attack this directly (and you really shouldn't), you would:

$\ \ \ $1) Find $P'(x)$.

$\ \ \ $2) Find the points in $[0,12]$ where $P'(x)=0$ or where $P'(x)$ is undefined.

$\ \ \ $3) Evaluate $P$ at the points found in 2) and at the points $0$ and $12$.

$\ \ \ $4) Select the $x$ value from 3) that gives the smallest pollution count.

Wolfram returns $x\approx8.77581$ as the distance from $A$ that gives the minimal amount of pollution.

Answer 2

Choose a co-ordinate system with Plant A as origin.
Basic trigonometry to find the angles of the triangle ABC
You will get (in Radians)
$\angle A = 0.9582$
$\angle B = 0.4214$
$\angle C = 1.762$

Now the position of Plant C is straightforward $(5\cos A, -5\sin A)$
Assuming the origin is fixed at Plant A the amount of pollution at any arbitary point (x, y) is $\frac{K}{x^2 + y^2 + 10}+ \frac{K}{4((x-12)^2 + y^2 + 10))} + \frac{K}{2((x-2.874968)^2+(y+4.0907895)^2+10))}$
On Road AB this reduces to $\frac{K}{x^2+10}+ \frac{K}{4((x-12)^2+10))} + \frac{K}{2((x-2.874968)^2+26.7345587))}$
Factoring out K and minimizing $\frac{1}{x^2+10}+ \frac{1}{4((x-12)^2+10))} + \frac{1}{2((x-2.874968)^2+26.7345587))}$
Minimum occurs at $x \approx 8.7758$ miles from Plant A on Road AB

Answer 3

My advice is to try to build the formula for the amount of pollution at an arbitrary point between plant A and plant B. With that formula, you can take the derivative and find where it's lowest.

As for how to find the formula, the amount of pollution you'd be exposed to would be the sum of the pollution from all three plants at any given moment.

Hope that helps.

Edit (made this into community wiki):

There seems to be some confusion about x. Suppose you're heading towards Plant B at mile 3. Pollution from plant A is therefore: $$\dfrac{K}{3^2 + 10}$$

Pollution from plant B then is: $$\dfrac{K}{4((12 - 3)^2 + 10)}$$

In order to know the pollution from plant C, we need to locate it with respect to its position between A and B.