1
$\begingroup$

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 (for some constant $K$):

$$\frac{K}{x^2 + 10}$$

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.

I am trying to find 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.

The pollution from Plant B must be

$$\frac{K}{4x^2 + 40},$$

and from Plant C must be

$$\frac{K}{2x^2 + 20}.$$

Adding the three equations would yield the following:

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

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

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

  • 0
    Ah nevermind, got it.2012-04-19
  • 0
    You've summed up the components wrong.2012-04-19
  • 0
    Ok... not sure how you got that, but I'll put that in and try to solve from there, thanks.2012-04-19
  • 0
    @Neil: Expression to be minimized is not right. Derivative isn't right either, but that's not important, since expression is the wrong one.2012-04-19
  • 0
    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 $K\over (12-x)^2+10$?2012-04-19
  • 0
    As Insidious shows, the distance from a given point is different for the three factories. So you can't use the same $x$ and add up the expressions. You need $x$ to be the position along the road and to express the pollution from each factory in terms of it.2012-04-19

3 Answers 3

0

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.

  • 0
    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?2012-04-19
  • 0
    I've updated my problem with how I'm trying to solve it.2012-04-19
  • 0
    @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.2012-04-19
  • 0
    X is the amount of miles from plants A, B, and C respectively.2012-04-19
  • 0
    @AndréNicolas Agreed, whoever gave this problem is a sadist.2012-04-19
  • 0
    @Archio, the problem with Plant C is that you're on the road from A to B, therefore you're not directly increasing or decreasing distance to C.2012-04-19
  • 0
    Isn't the distance to C increasing/decreasing as you move along Road A-B? How do I locate pollution to plant C within it's respective position?2012-04-19
  • 0
    Distance increases as you approach the road to plant C and decreases afterwards. You need to figure out the position of Plant C given the information provided.2012-04-19
  • 0
    If you are on the road from $A$ to $B$, $x$ miles from $A$, I calculate the distance to $C$ to be $\sqrt{x^2+{69\over12}x+25}$.2012-04-19
  • 0
    I see, that makes sense... so I need to somehow find the derivative of that, which will give me the lowest point of pollution?2012-04-19
  • 0
    @Archio Yes you'd take $f(x)=$ (poll. from A)+(poll. from B)+(poll. from C). Then minimize $f$ over $[0,12]$. You'll have to find $f'$ and then solve $f'(x)=0$. Then determine the minimum. This will not be easy... Where did you get this problem?2012-04-19
  • 0
    Thanks, I'm going to try that. I got it in my Calculus I class, believe it or not.2012-04-19
1

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

  • 0
    I don't think the third term in your reduced equation is correct. $y=0$, but what happened to the $4.0907^2$?2012-04-19
  • 0
    @DavidMitra Thanks, fixed it!2012-04-20
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.

  • 0
    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.2012-04-19
  • 0
    @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 $P_C(x)$).2012-04-19