Two towns A and B are to get their water supply from the same pumping station to be located on the bank of a straight river that is 15 km from town A and 10 km from town B. If the points on the river nearest to A and B are 20 km apart and A and B are on the same side of the river, where should the pumping station be located so that the least amount of piping is required.
Let M be location and L be amount of piping
L= AM + BM
Found AM^2 and BM^2
Stuck after that
Assuming that the pipes are to be straight from M to A and B (that is, the pipe does not go to some other point P and then divide), then let's add a couple more points to the diagram:
P: the point on the river bank closest to A
Q: the point on the river bank closest to B
Then PM+MQ = √(20^2-5^2) = √395
Then if M is at distance x from P, it is √395-x from Q.
The amount of piping is AM+BM where
AM^2 = x^2+15^2
BM^2 = (√395-x)^2 + 10^2
So, we have
L = √(x^2+225) + √(x^2-2x√395+495)
L is a minimum when x = 3√(79/5) = 11.92
Not sure what analytical tools you have at your disposal to solve this kind of problem.
Thanks so much. May I know what are the in-between steps for the following two lines please?
L = √(x^2+225) + √(x^2-2x√395+495)
L is a minimum when x = 3√(79/5) = 11.92
Ah, the great water distribution dilemma! Well, let me dive right into it and give you a funny answer.
Now, since we want to minimize the amount of piping required, we need to find the sweet spot for our pumping station. Let's imagine the river as a giant snake, slithering between town A and town B. Our goal is to tickle the snake as little as possible with our piping.
First, let's look at town A. It's like a lonely person standing 15 km away from the riverbank, trying to quench its thirst. Poor A! Meanwhile, town B is a bit closer, only 10 km away. So, to gather the required H2O, we need to make a decision.
If we place the pumping station closer to A, we can minimize the piping distance from A to the station, but then we would have to deal with more piping from the station to B. On the other hand, if we place it closer to B, the piping distance from B to the station will be shorter, but then we'll have more piping from the station to A.
So, where should the pumping station be located? I suggest finding a place somewhere in the middle, where both A and B can appreciate the beauty of the river equally. That way, we can avoid unnecessary pipe bending and keep the piping distance balanced. After all, we don't want to stress out the snakes!
To find the location of the pumping station that requires the least amount of piping, you can use the principle of minimum distance.
Let's say the distance from the pumping station to town A is 'x' km, and the distance from the pumping station to town B is 'y' km.
According to the given information, the points on the river nearest to town A and B are 20 km apart, and the river is located 15 km from town A and 10 km from town B.
From this information, we can form the following equations:
1. The distance between the pumping station and town A (AM) + the distance between the pumping station and town B (BM) equals the distance between town A and town B (AB):
AM + BM = AB
2. The sum of the distances from town A and town B to the river equals the distance between the points on the river nearest to them:
AM + 15 = BM + 10
AM - BM = -5
Now, we can solve these equations simultaneously to find the values of x and y, which will give us the proximity of the pumping station to each town:
AM + BM = AB (Equation 1)
AM - BM = -5 (Equation 2)
By adding Equation 1 and Equation 2, we have:
2AM = AB - 5
Substituting the value of AB (distance between the points on the river nearest to town A and B), which is 20 km:
2AM = 20 - 5
2AM = 15
AM = 7.5
Now, substitute the value of AM in Equation 2:
7.5 - BM = -5
BM = 7.5 + 5
BM = 12.5
Therefore, the pumping station should be located 7.5 km from town A and 12.5 km from town B to require the least amount of piping.
To find the location of the pumping station where the least amount of piping is required, you need to minimize the sum of the distances from the pumping station to both towns.
Let's break down the problem step by step:
1. Visualize the problem:
Draw a diagram with the river as a straight line. Label the two towns A and B on the same side of the river. Mark the locations where the river is nearest to each town and label them as C and D. Also, mark the location of the pumping station as M.
The given information tells us that AC = 15 km, BD = 10 km, and CD = 20 km.
A C M D B
|----|----------|----------|----|
15 km ? 10 km
2. Set up the problem:
We want to minimize the total amount of piping required. Let's define L as the total piping needed. Since the pumping station is located at M, we can derive the equation: L = AM + BM.
3. Break down the distances:
Consider triangle AMC. By applying the Pythagorean theorem, we can find that AM^2 = AC^2 − CM^2.
Similarly, consider triangle BMD. By applying the Pythagorean theorem, we can find that BM^2 = BD^2 − DM^2.
We are given AC and BD from the problem statement, but we need to find CM and DM.
4. Use geometry to find CM and DM:
In triangle CMD, we have two sides: CM = CD − DM and DM = CD − CM. Substituting these expressions into the equations from step 3, we get:
AM^2 = AC^2 − (CD − DM)^2 (equation 1)
BM^2 = BD^2 − (CD − CM)^2 (equation 2)
Notice that the equations involve both CM and DM. To solve this, we need to eliminate one of the variables.
5. Use the property of similar triangles:
Observe that triangles ACM and BDM are similar, as they share an angle at C and D and have right angles at M. Therefore, their sides are proportional:
AC/AM = DC/DM (equation 3)
BD/BM = CD/CM (equation 4)
Rearrange equation 3 to solve for CM:
CM = (CD * AM) / AC (equation 5)
Rearrange equation 4 to solve for DM:
DM = (CD * BM) / BD (equation 6)
6. Substitute CM and DM in equations 1 and 2:
Substitute equations 5 and 6 into equations 1 and 2:
AM^2 = AC^2 − [(CD * BM) / BD]^2
BM^2 = BD^2 − [(CD * AM) / AC]^2
7. Simplify the equations and work towards minimizing L:
Expand the equations and simplify:
AM^2 = AC^2 − (CD^2 * BM^2) / BD^2
BM^2 = BD^2 − (CD^2 * AM^2) / AC^2
We now have two equations involving AM^2 and BM^2 in terms of known distances.
8. Minimize L:
To minimize L = AM + BM, we need to minimize AM^2 + BM^2. We established the equations in step 7. Solve them simultaneously to find the values of AM and BM.
Once you find the values, you can calculate L = AM + BM. This will give you the least amount of piping required.
Remember, in this explanation, I have provided the steps to solve the problem analytically. You can solve the equations using algebra or numerical methods to find the values of AM, BM, and L.