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

enter image description here

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.

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

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Ah nevermind, got it. –  Neil Apr 19 '12 at 14:51
    
You've summed up the components wrong. –  Neil Apr 19 '12 at 15:17
    
Ok... not sure how you got that, but I'll put that in and try to solve from there, thanks. –  Archio Apr 19 '12 at 15:20
    
@Neil: Expression to be minimized is not right. Derivative isn't right either, but that's not important, since expression is the wrong one. –  André Nicolas Apr 19 '12 at 15:22
    
This seems vaguely stated to me. If you are on the road from $A$ to $B$ and $x$ miles from $A$, then you are $12-x$ miles from $B$. Is the pollution from $B$ at that point $1/4$ of $K\over x^2+10$ or $1/4$ of $K\over (12-x)^2+10$? –  David Mitra Apr 19 '12 at 15:22

3 Answers 3

up vote 1 down vote accepted

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.

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I'm confused- which is the right solution, yours or the one Insidious posted? Thank you so much for helping, but I don't know which answer to refer to. –  Archio Apr 19 '12 at 20:31
    
@Archio We have the same solution, though I kept mine exact. I believe Insidious made a typo going from his first displayed equation to his second. The third term in his second equation should have a $(4.0907895)^2$ in it (which corresponds to my $1071/64$ under the square root in the expression for $P_C(x)$). –  David Mitra Apr 19 '12 at 20:53

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

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I don't think the third term in your reduced equation is correct. $y=0$, but what happened to the $4.0907^2$? –  David Mitra Apr 19 '12 at 17:22
    
@DavidMitra Thanks, fixed it! –  Insidi0us Apr 20 '12 at 4:02

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.

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I'm not sure I understand how to find the formula... if I add all of the equations, how does K factor in? Will the measures be on Road A-B? –  Archio Apr 19 '12 at 14:57
    
I've updated my problem with how I'm trying to solve it. –  Archio Apr 19 '12 at 15:05
    
@Archio: You do not say what $x$ is. Assume that it is distance to $A$. Pollution from $A$ is $\frac{K}{x^2+10}$. But distance from $B$ is $12-x$, so pollution from $B$ is $\frac{K/4}{(12-x)^2+10}$. Pollution from $C$ is a nightmare. We probably need to find distance of $C$ from road. I did it, it is ugly. Then you can find pollution from $C$ at the point "$x$." Real ugly. Add up, differentiate. It will be horrible. –  André Nicolas Apr 19 '12 at 15:19
    
X is the amount of miles from plants A, B, and C respectively. –  Archio Apr 19 '12 at 15:21
    
@AndréNicolas Agreed, whoever gave this problem is a sadist. –  David Mitra Apr 19 '12 at 15:25

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