Finding Distance in Plan 1,2,3 & 4: Minimizing Cost

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In summary, the conversation discusses various diagrams and their corresponding distances and angles between points P, Q, and R. The goal is to find the shortest distance or "plan" in order to minimize cost, assuming that the cost per kilometer is the same. The distances for each plan are: 5 km for Plan 1, 6 km for Plan 2, 3/(2sqrt(2)) km for Plan 3, and (3+ sqrt(6))/sqrt(3) km for Plan 4. The plan with the shortest distance will also have the minimum cost.
  • #1
dagg3r
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P is 2km from Q and R lies on the perpendicular bisector at the link between P and Q and distance from P to R is 3 km.


1. Find Length for Plan 1
2. " " " Plan 2
3. " " " Plan 3
4. " " " Plan 4

5. Decide which plan would be used if minimizing the cost.

http://www.jeack.com.au/~dngo/maths.bmp

are the diagrams.

i solved question 1 + 2.

just need help from 3 - 5.

if some people do not wish to look at the picture i will give a visual description in words.

diagram 3 = is in a Y shape. the distance in the middle from P to Q is x. and the stroke down which is R, opposite sides of R is x. so altogether there are 3 unknowns.

diagram 4 = is in a Y shape, P to Q is equal which has sides b. the distance in the middle from P to Q is 120. opposites of R is 120.

i suggest you can please look at the web link

http://www.jeack.com.au/~dngo/maths.bmp thanks.
 
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  • #2
I looked at the diagrams and they are very clear. However, I still don't know what the "cost" is because you said nothing about what these diagrams are or how money comes into it!

Are these "pipes" or "roads" between areas? If so what is the cost of building km of it?

If the diagrams are of some sort of connection to be built connecting P, Q, and R, and the cost "per km" is the same everywhere, then you are really asking for the distances.

Picture 1 shows a connection directly from Q to P (distance 2 km) and then a connection from P to R (distance 3 km) so the total distance is 5 km.

Picture 1 shows a connection from Q to R and then from P to R. We are told that the distance from P to R is 3 km. Since R is on the perpendicular bisector of the segment from P to Q, by symmetry, the distance from P to R is also 3 km. Total distance: 6 km.

Picture 3 has links from P,Q, and R to a central point. We are told that the 3 distances must be the same (x). This may not be the simplest way to do it, but I set it up on a coordinate system. Take the x-axis to be the line through P and Q. Since P is on the perpendicular bisector of the segment through P and Q we can take P on the y-axis, P as (-1,0) and Q as (1,0). Then, if call P (0,p), the distance from P to Q is given by sqrt(1+ p^2)= 3. Squaring both sides, 1+p^2= 3 or p^2= 2 so p= sqrt(2). The "central point" we are looking for in (3) must also (again by symmetry) be on the y axis. Call it S= (0,s). Now we have: distance from P to S is sqrt(1+s^2) and that must be the same as the distance from R to S: sqrt(2)- s (Since they are both on the y-axis, we just subtract the y coordinates. Strictly speaking we should take absolute value but it is clear from the picture that s is less than sqrt(2).) The two distances are the same: sqrt(1+ s^2)= sqrt(2)-s. Squaring both sides
1+ s^2= 2- 2sqrt(2)s+ s^2 or 2sqrt(2)s= 1 so s= 1/(2sqrt(2)). The total distance is 3s= 3/(2sqrt(2)).

In (4) you have either misunderstood your own picture or miswritten your description. "120" is obviously not a "distance": it is the angle in degrees between each segment. (4) is the same as (3) except that instead of requiring each segment be of the same length we are requiring that each leg be the same angle from the others.
Again, call the central point "S". Setting up a coordinate system as before we can see (again, from symmetry) that S is on the y-axis: S= (0,s). The angle between the segment from P to S and the segment from R to S is 120 degrees so the angle between the line from P to S and the horizontal is 120- 90= 30 degrees. Since PQ and that horizontal are parallel the angle between PQ and PR is also 30 degrees (alternating interior angles). The distance from P to the point bisecting PQ is 1 so s/1= tan 30= 1/sqrt(3). S is the point
(0, 1/sqrt(3)). From that it is easy to calculate that PS= sqrt(1+ 1/3)= 2/sqrt(3). This is, by symmetry, the same as QS. Since R is still the point (0,sqrt(2)), the distance from R to S, RS, is
sqrt(2)- 1/sqrt(3)= (sqrt(6)-1)/sqrt(3). The total distance, then, is
2/sqrt(3)+ 2/sqrt(3)+ (sqrt(6)-1)/sqrt(3)= (3+ sqrt(6))/sqrt(3).

(5) asks for the "plan" which will minimize cost. Assuming that the cost of each km is the same (If different areas require different costs, that would change everyhing!) then the "minimum cost" is the same as minimum distance. The distances are:
(1) 5 km.
(2) 6 km.
(3) 3/(2sqrt(2)) km.
(4) (3+ sqrt(6))/sqrt(3) km.

I'll let you calculate those yourself and see which is smallest.
 
  • #3


To find the length for Plan 3, we can use the Pythagorean theorem to solve for the unknown sides. From the diagram, we can see that the distance from P to R is 3km, and the distance from Q to R is also 3km. Therefore, we can set up the following equation:

x^2 + 3^2 = (x+3)^2

Expanding the right side and simplifying, we get:

x^2 + 9 = x^2 + 6x + 9

Subtracting x^2 and 9 from both sides, we get:

6x = 0

Dividing both sides by 6, we get:

x = 0

Therefore, the length for Plan 3 would be 0km.

For Plan 4, we can use the same approach. From the diagram, we can see that the distance from P to Q is b, and the distance from P to R is 120km. Therefore, we can set up the following equation:

b^2 + 120^2 = (b+120)^2

Expanding the right side and simplifying, we get:

b^2 + 14400 = b^2 + 240b + 14400

Subtracting b^2 and 14400 from both sides, we get:

240b = 0

Dividing both sides by 240, we get:

b = 0

Therefore, the length for Plan 4 would also be 0km.

To decide which plan would be used to minimize cost, we need to consider the cost of each plan. Since Plan 3 and Plan 4 both have a length of 0km, the cost for these plans would also be 0. However, for Plan 1 and Plan 2, the cost would be based on the length of the link between P and Q. Since we do not have enough information about the length of this link, we cannot determine the cost for these plans. Therefore, we cannot make a decision on which plan would be used to minimize cost.
 

1. How do I calculate the distance in Plan 1, 2, 3, & 4?

To calculate the distance in these plans, you will need to know the coordinates of the starting point and the destination. Then, you can use the Pythagorean theorem (a^2 + b^2 = c^2) to find the distance, where a and b are the differences in the x and y coordinates, and c is the distance. Repeat this process for each plan to find the distances in all four plans.

2. Why is it important to minimize cost in finding distance?

Minimizing cost is important because it helps to optimize resources and save time. By finding the most efficient route or plan, you can reduce travel time and expenses, leading to increased productivity and cost savings. This can be particularly beneficial in industries such as transportation, logistics, and supply chain management.

3. How can I use Plan 1, 2, 3, & 4 to minimize my travel costs?

You can use Plan 1, 2, 3, & 4 to minimize your travel costs by comparing the distances in each plan and choosing the one with the shortest distance. This will help you to save on transportation costs and potentially reduce any additional costs, such as toll fees or fuel expenses.

4. Can I use Plan 1, 2, 3, & 4 to find the shortest distance between multiple points?

Yes, you can use these plans to find the shortest distance between multiple points. Simply calculate the distances between each point and choose the plan with the overall shortest distance, or combine different plans to create a more efficient route.

5. Are there any limitations to using Plan 1, 2, 3, & 4 to minimize cost?

There may be limitations to using these plans to minimize cost, depending on the specific situation. For example, these plans may not take into account factors such as traffic, road conditions, or specific route restrictions. It is important to consider all relevant factors and use your judgement when choosing the best plan to minimize cost.

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