An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6 km east of the refinery. The cost of laying pipe is $200,000/km over land to a point P on the north bank and $400,000/km under the river to the tanks. To minimize the cost of the pipeline, how far (in km) from the refinery should P be located? (Round your answer to two decimal places.)
Well, it seems like we have a puzzling pipeline predicament!
To minimize the cost of the pipeline, we need to find the best spot for point P on the north bank. Now, let's do some math comedy!
Since the river is 2 km wide, we can imagine it as a big swimming pool for the pipeline. If we choose P too far to the east, then the pipeline will have to swim a long way across the river. That'll cost us big bucks!
On the other hand, if we choose P too close to the refinery, the pipeline will need to travel a long distance over the land. That's like forcing it to walk on hot coals – it'll also cost us a pretty penny!
So we need to find the sweet spot where the pipeline swims just enough to minimize the cost. Let's call that sweet spot x km from the refinery.
Now, the cost of the underwater pipeline is $400,000 per km, but the cost of the overland pipeline is only $200,000 per km. We want to minimize the total cost, so we need to balance the distance over land and the distance underwater.
If we choose P to be x km from the refinery, then the distance over land is 6 - x km, and the distance underwater is 2 km.
So the total cost will be (6 - x) * $200,000 + 2 * $400,000 = $1,200,000 - $200,000x + $800,000
To minimize this cost, we need to find the value of x that minimizes the expression $1,200,000 - $200,000x + $800,000.
To find that value, we can take the derivative of the expression with respect to x and set it equal to zero. But you know what? Let's not dive into those scary calculus waters!
Instead, we'll use our comedy intuition to realize that the expression is a linear function, which means the minimum value occurs at one of the endpoints. In this case, the only endpoints we have are x = 0 and x = 6.
Plugging these values into the expression, we find that the cost at x = 0 is $2,000,000, and the cost at x = 6 is $2,200,000.
So, the minimum cost occurs when we choose P to be right at the refinery (x = 0)! That saves us $2,000,000, and we can use that money to throw a circus-themed party. Clown noses for everyone!
Therefore, to minimize the cost of the pipeline, point P should be located 0 km from the refinery. Keep the clown comedy flowing and the pipeline cost low!
To minimize the cost of the pipeline, we need to find the point P on the north bank that minimizes the total cost of laying the pipe.
Let's assume that the distance from the refinery to point P is x kilometers. Therefore, the distance from point P to the storage tanks on the south bank would be (2 - x) kilometers.
The cost of laying pipe over land from the refinery to point P would be $200,000 times x (total cost 1), and the cost of laying pipe under the river from point P to the storage tanks would be $400,000 times (2 - x) (total cost 2).
The total cost of the pipeline would be the sum of total cost 1 and total cost 2, which is:
Total cost = (200,000 * x) + (400,000 * (2 - x))
To minimize this cost, we can find the derivative of the total cost with respect to x and set it equal to zero.
d(Total cost) / dx = 200,000 - 400,000
Setting the derivative equal to zero, we have:
200,000 - 400,000 = 0
-200,000 = 0
This equation has no solution, which means that there is no minimum point for the cost function. Therefore, the cost will either continuously increase or continuously decrease as x varies from 0 to 2.
To identify the best location for point P, we can compare the cost at the extremes of the range:
1. When x = 0 (directly at the refinery), the total cost would be: (200,000 * 0) + (400,000 * 2) = $800,000
2. When x = 2 (directly across from the storage tanks), the total cost would be: (200,000 * 2) + (400,000 * 0) = $400,000
Based on these calculations, we can conclude that locating point P directly across from the storage tanks (at x = 2) results in the minimum cost for the pipeline.
To minimize the cost of the pipeline, we need to find the optimum distance for point P on the north bank. Let's analyze the problem step by step:
1. Draw a diagram: Draw a diagram representing the situation described in the problem. Draw the river, the refinery on the north bank, and storage tanks on the south bank. Label the width of the river as 2 km and the distance from the refinery to the storage tanks as 6 km.
2. Assign variables: Let's assign variables to the distances we need to find. Let's denote the distance from the refinery to point P on the north bank as x km.
3. Calculate the cost of overland pipe: The cost of laying the pipe overland from the refinery to point P can be calculated by multiplying the overland cost per kilometer ($200,000/km) by the distance x km.
4. Calculate the cost of underwater pipe: The cost of laying the pipe underwater from point P to the storage tanks can be calculated by multiplying the underwater cost per kilometer ($400,000/km) by the distance across the river, which is 2 km.
5. Calculate the total cost: To minimize the cost, we need to minimize the total cost. The total cost is the sum of the cost of the overland pipe and the cost of the underwater pipe.
Total Cost = Overland Cost + Underwater Cost
6. Express the total cost in terms of x: Using the calculations from steps 3 and 4, we can express the total cost in terms of x.
Total Cost = (200,000 * x) + (400,000 * 2)
7. Simplify the expression: Simplify the expression for the total cost.
Total Cost = 200,000x + 800,000
8. Find the minimum point: To find the minimum cost, we need to find the minimum of the total cost expression. This occurs at the minimum point on the graph of the expression. Since the expression represents a linear function (straight line), the minimum will be at the endpoint of the line segment.
9. Determine the distance for minimum cost: The distance for the minimum cost corresponds to the x-coordinate at the endpoint of the line segment. To find this, we need to identify which endpoint is closer to the refinery.
- If the endpoint on the north bank is closer, point P will be located at x = 0 km (right at the refinery).
- If the endpoint on the south bank is closer, point P will be located at x = 6 km (right at the storage tanks).
10. Calculate the optimized distance: Compare the distances identified in step 9 to find the x-coordinate that results in the minimum cost. Round this value to two decimal places.
In this case, the distance from the refinery to point P on the north bank should be 0 km, which means point P should be located right at the refinery. This will minimize the cost of the pipeline.
As usual, draw a diagram. Let
R = refinery
T = tank farm
Q = point directly across the river from T
x = the distance RP
So QT = 2
The land distance is x
underwater distance is z^2 = (6-x)^2 + 2^2
Now the cost (in $100,000s) of the pipeline is
c(x) = 2x + 4√((6-x)^2 + 4)
now just find where dc/dx = 0
x = 6 - 2/√3
Note that as usual in these problems, the angle θ made by the pipeline with the coastline is such that sinθ is the ratio of the cost on land to that in water
In this case, tanθ = 2/4 = 1/2. You note that PQ/QT = 1//3
My sketch has R as the refinery on the north side, S as the location of the
storage tanks on the south side, and Q as the point directly across the river from S. Let PQ = x, then PR = 6-x, and QS = 2
Path of pipeline = SP + PR
Since we are not asked for the actual cost, only the path route, and the fact
that I noticed that the costs of laying the pipe are simply in the ratio of 2 : 1,
we can say:
Cost = 2SP + 1PR
also SP^2 = x^2 + 2^2
we have
Cost = 2(x^2 + 4)^(1/2) + 1(6-x)
dCost/dx = (x^2 + 4)^(-1/2)(2x) - 1 = 0 for a min of Cost
2x/√(x^2 + 4) = 1
2x = √(x^2 + 4)
square both sides
4x^2 = x^2 + 4
3x^2 = 4
x = 2/√3 =1.16 km
so 6-x = 4.85
They should aim for a point 4.85 km from the refinery