Calculus 3 prob: optimal locations for the city's new zoo and new hospital.

[mrITD]

A city planning commission is seeking to identify prime locations for a new zoo and a new hospital. The city's economy is heavily dependent on two industrial plants located relatively near the city center. Both emit particulate matter into the atmosphere, and the resulting air pollution is of concern to the commission and will influence its decisions. The consensus of the commission is that the new zoo should be built in the least polluted area within the city limits, and the new hospital should be built in the least polluted location within two miles of the city center.

The boundaries of the city form a rectangle ten miles from east to west and six miles from north to south. When a coordinate system is chosen with the origin at the center of the rectangle (the city center), Industrial Plant 1 has coordinates (-1, 1) and Industrial Plant 2 has coordinates (1, 0). At a point, (x, y), the concentration of particulate matter (in parts per million) due to emissions from Plant 1 is given by C₁ = 200 - 3(d₁)², where d₁ is the distance from (x, y) to Plant 1. Similarly, the concentration due to Plant 2 is given by C₂ = 200 - 3(d₂)², where d₂ is the distance from (x, y) to Plant 2.

(A) Find the point within the city limits that has the greatest concentration of particulate matter.

(B) Find the points on the city boundary that have the greatest and least concentrations of particulate matter.

(C) Find the average concentration of particulate matter throughout the city.

(D) Find the points on the circle of radius two miles, centered at the origin, that have the greatest and least concentrations of particulate matter.

(E) Determine the optimal locations for the city's new zoo and new hospital.

Help what can i do?

 

[stapel]

A city planning commission is seeking to identify prime locations for a new zoo and a new hospital. The city's economy is heavily dependent on two industrial plants located relatively near the city center. Both emit particulate matter into the atmosphere, and the resulting air pollution is of concern to the commission and will influence its decisions. The consensus of the commission is that the new zoo should be built in the least polluted area within the city limits, and the new hospital should be built in the least polluted location within two miles of the city center.

The boundaries of the city form a rectangle ten miles from east to west and six miles from north to south. When a coordinate system is chosen with the origin at the center of the rectangle (the city center), Industrial Plant 1 has coordinates (-1, 1) and Industrial Plant 2 has coordinates (1, 0). At a point, (x, y), the concentration of particulate matter (in parts per million) due to emissions from Plant 1 is given by C₁ = 200 - 3(d₁)², where d₁ is the distance from (x, y) to Plant 1. Similarly, the concentration due to Plant 2 is given by C₂ = 200 - 3(d₂)², where d₂ is the distance from (x, y) to Plant 2.

(A) Find the point within the city limits that has the greatest concentration of particulate matter.

(B) Find the points on the city boundary that have the greatest and least concentrations of particulate matter.

(C) Find the average concentration of particulate matter throughout the city.

(D) Find the points on the circle of radius two miles, centered at the origin, that have the greatest and least concentrations of particulate matter.

(E) Determine the optimal locations for the city's new zoo and new hospital.

Help what can i do?

A good place to start would probably be with the algebra. What can you do with the provided coordinate system, generic point (x, y), the two distance functions, and the Distance Formula?

 

[mrITD]

I got :
C1(X,Y) = 200-3(x^2+y^2)
C 2(X,Y) = 200-3(X^2+Y^2)
now what can i do? be specific at this part please.

 

[stapel]

I got :
C1(X,Y) = 200-3(x^2+y^2)
C 2(X,Y) = 200-3(X^2+Y^2)

How did you get this? Why are you not using the Distance Formula? What is your specific reasoning for whatever you are using instead?

Please be complete. (I'm glad to work this doing it your way, rather than what you were given before, but I'll need first to understand what you're doing, and currently I don't.) Thank you! :wink:

 

[mrITD]

How did you get this? Why are you not using the Distance Formula? What is your specific reasoning for whatever you are using instead?

Please be complete. (I'm glad to work this doing it your way, rather than what you were given before, but I'll need first to understand what you're doing, and currently I don't.) Thank you! :wink:

I saw the distance formula but in that examples they are calculating the distances of the points that we already know them , but in this exercise says "where d1 is the distance from (x, y) to Plant 1. Similarly, the concentration due to Plant 2 is given by C2 = 200 - 3(d2)2, where d2 is the distance from (x, y) to Plant 2."


So i need to known how much is d1 , so I used [the Pythagorean Theorem in order to create an expression for ]the distance from (x, y) to (-1, 1).


i got that d1 =square root(x^2+y^2) and d2 = square root(x^2+y^2)


i substitute that value in


C1 = 200 - 3(d1)2
C2 = 200 - 3(d2)2


and i got the result that i mentioned before.


I think that i´m wrong so..


can you give me the value of d1 and d2? also i want to know your way, i´m stuck in this problem :/

 

[stapel]

... need to known how much is d₁ , so used [the Pythagorean Theorem] in [order] to [create an expression for ]the distance from (x, y) to (-1, 1).

Ah. that makes much more sense. Thank you.

got that d₁ = square root(x^2 + y^2) and d₂ = square root(x^2 + y^2)

How do these take the points (-1, 1) and (1, 0) into account?

Let's try using the information they gave you. You have two plants, located at (-1, 1) and (1, 0), respectively. You have generic locations, (x, y). You have the Distance Formula, which provides a way of find expressions for the distance between (-1, 1) and (x, y) and the distance between (1, 0) and (x, y). Start by plugging the three given points into the Distance Formula, and thus finding expressions for each of the two distances. Then plug these into the "concentration" equations.

Since you are wanting to minimize the total concentrations, it might be helpful then to consider adding the two "concentration" expressions (being the right-hand sides of the provided "concentration" equations). Then work from there.

 

[Mullins]

?

Hello,

I saw this problem and liked it enough to try solving it, however I have encountered a problem that could well lead to a different thread.

Anyway, I have 2 ideas:

First - naive:
Just drawing the picture and looking at the equations we can see that the pollution has to be highest in between the plants, specifically lying on the middle of the line running through them, since the pollution equations are the same.
By the same logic we could see that the Hospital will lie on the circle (meaning 2 miles from origin) and the distance would be longest possible from both plants (then the d₁=d₂). Similarly this would apply to the ZOO location just on the borders of city.
However somehow I would have to solve the exact points and it would not be very helpful for solving the other questions.

Second - less naive:
Using method of Lagrange multiplicators for bounded extrema, we could plug in the C₁ and C₂ (of course using the distance formula instead of d₁ and d₂). Then the boundary would be M₁={(x,y) in R² | x² + y² = 4} for the Hospital and M₂={(x,y) in R² | (x less or equal |5|) and (y less or equal |3|)} for the ZOO. Then taking derivatives and solving should be quite straightforward.

But how do I transform the set M₂ as it is currently written into something workable? Namely I would like an equation in form ax + by + c = 0 or something like that, which could be easily plugged in as the boundary condition. But I don't know how to get it - the only idea for getting rid of the absolute values is to square them, and since we are only interested in the points lying on the rectangle, we could get rid of the inequalities. So now we have 2 boundary conditions x² - 25 = 0 and y² - 9 = 0. That should be okay with the Lagrange multiplicators method.

However, what would I do in the case that the entire area within the city borders would be of interest? How would I define my boundary conditions?

Thank you in advance.

 

[Mullins]

In the meantime...

Hello again,

so I actually managed to solve the problem myself, and would thus answer myself a little bit.

First - naive:
Just drawing the picture and looking at the equations we can see that the pollution has to be highest in between the plants, specifically lying on the middle of the line running through them, since the pollution equations are the same.
By the same logic we could see that the Hospital will lie on the circle (meaning 2 miles from origin) and the distance would be longest possible from both plants (then the d₁=d₂). Similarly this would apply to the ZOO location just on the borders of city.
However somehow I would have to solve the exact points and it would not be very helpful for solving the other questions.

So this is just a way for discovering the solution to A), where it actually works, however it is not a good approach. The easiest is grad C(x,y) = 0.

Second - less naive:
Using method of Lagrange multiplicators for bounded extrema, we could plug in the C₁ and C₂ (of course using the distance formula instead of d₁ and d₂). Then the boundary would be M₁={(x,y) in R² | x² + y² = 4} for the Hospital and M₂={(x,y) in R² | (x less or equal |5|) and (y less or equal |3|)} for the ZOO. Then taking derivatives and solving should be quite straightforward.

But how do I transform the set M₂ as it is currently written into something workable? Namely I would like an equation in form ax + by + c = 0 or something like that, which could be easily plugged in as the boundary condition. But I don't know how to get it - the only idea for getting rid of the absolute values is to square them, and since we are only interested in the points lying on the rectangle, we could get rid of the inequalities. So now we have 2 boundary conditions x² - 25 = 0 and y² - 9 = 0. That should be okay with the Lagrange multiplicators method.

This is where I was wrong in the case of ZOO. The set M₂ as I defined it is actually just the 4 vertices on the city borders, which doesn't help me a lot. When I discovered that, I thought that taking just one of the conditions at a time would help me, but I didn't get a reasonable answer (for a minimum), because there is no minimum, the curve just keeps going down to negative infinity as there is no boundary.

However, what would I do in the case that the entire area within the city borders would be of interest? How would I define my boundary conditions?

It turned out that the method of Lagrange multipliers is not fit to deal with inequalities and the way to go is as follows (thanks to Christian Blatter: http://math.stackexchange.com/a/49815):

Solving Lagrange multipliers for the edges one condition at a time as I mentioned before is the first step. We get points of interest on the edges E_i, add the vertices V_i and, if we are interested in the area within borders, solve grad C(x,y)=0 and add all those points A_i. From this set L={A_i,E_i,V_i}, we take either minimum or maximum, depending on what we are interested in.