A gas pipeline is to be constructed from a storage tank, which is right on a road, to a house which is 600 feet down the road and 300 feet back from the road. Pipe laid along the road costs $8.00 per foot, while pipe laid off the road costs $10.00 per foot. What is the minimum cost for which this pipline can be built?
i don't get how to get x =200 when i set to 0?
To determine the minimum cost for building the pipeline, we need to find the point where the cost is minimized. Let's break down the problem step by step:
1. Define the variables:
- x: length of pipeline laid along the road (in feet)
- y: length of pipeline laid off the road (in feet)
2. Set up an equation based on the given information:
- The total length of the pipeline is x + y (the sum of the lengths laid along the road and off the road).
- The cost of the pipeline is calculated as follows:
- Cost along the road = 8 * x (since it costs $8.00 per foot along the road)
- Cost off the road = 10 * y (since it costs $10.00 per foot off the road)
- Total Cost = Cost along the road + Cost off the road = 8x + 10y
3. Determine the constraints of the problem:
- The house is located 600 feet down the road and 300 feet back from the road, forming a right-angled triangle.
- Using the Pythagorean theorem, we have: x^2 + y^2 = (600^2 + 300^2) (since the hypotenuse is the length of the pipeline).
4. Solve for x in terms of y in the constraint equation:
- Start with: x^2 + y^2 = 600^2 + 300^2
- Rearrange the equation: x^2 = (600^2 + 300^2) - y^2
- Take the square root of both sides: x = √[(600^2 + 300^2) - y^2]
5. Substitute the value of x into the total cost equation:
- Total Cost = 8x + 10y
- Total Cost = 8√[(600^2 + 300^2) - y^2] + 10y
6. To find the minimum cost, we need to find the value of y that minimizes the Total Cost:
- Differentiate the Total Cost equation with respect to y and set it equal to zero:
- d(Total Cost) / dy = 0
- Solve for y to find the value that minimizes the cost.
7. Once you solve for y, substitute it back into the Total Cost equation to find the minimum cost.
It seems that you mentioned x = 200 when you set it to zero. However, without further information or context, it is unclear how you arrived at that value.
To calculate the minimum cost for constructing the gas pipeline, we need to determine the optimal distance to lay the pipe along the road and off the road. Let's break down the problem step by step.
First, let's define the variables:
- Let x be the distance (in feet) at which the pipeline is laid along the road.
- Let y be the distance (in feet) at which the pipeline is laid off the road.
We know that:
- The distance from the storage tank to the house is 600 feet down the road.
- The distance from the house to the road is 300 feet back.
To find the minimum cost, we need to set up an equation that represents the total cost of laying the pipeline.
Since the cost of laying pipe along the road is $8.00 per foot, the cost for that section is 8x.
Similarly, the cost of laying the pipe off the road is $10.00 per foot, so the cost for that section is 10y.
The total cost equation becomes:
Total Cost = Cost along the road + Cost off the road
Total Cost = 8x + 10y
Now let's take the constraints into consideration.
The pipeline needs to reach the house, so the total distance traveled along the road and off the road should sum up to 600 feet:
x + y = 600
The pipeline needs to travel 300 feet back from the road to reach the house, so we have another constraint:
y = 300
Now we have a system of two equations:
x + y = 600
y = 300
To solve this system of equations, we can substitute the value of y from the second equation into the first equation:
x + 300 = 600
x = 600 - 300
x = 300
Therefore, x = 300 feet.
So, the optimal distance to lay the pipeline along the road is 300 feet.
Regarding your question about how to get x = 200 when you set it to 0, it seems to be a misunderstanding or a different problem. In this particular scenario, we found that x = 300 feet, not 200, using the necessary constraints and equations.
My diagram:
A --- storage tank
B --- point on road, 600 ft down from A
Q ---- point we want to go to, AB perpendicular to BQ
P ---- point between A and B where we want to leave the road.
Let PB = x
so we have:
PB = x, AP = 600-x, BQ = 300
L(ength) of pipeline = AP + PQ = (600-x) + √(x^2 + 300^2)
cost = 8(600-x) + 10(x^2 + 90000)^(1/2)
d cost/dx = -8 + 5(2x)(x^2 + 90000)^(-1/2) = 0 for a min of cost
10x / √(x^2 + 90000) = 8
5x = 4√(x^2 + 90000)
square both sides:
25x^2 = 16x^2 + 360000
9x^2 = 360000
x^2 = 40000
x = 200