To solve this, we will minimize the total length of the pipelines from both manufacturing plants to the lake. Let’s use mathematical optimization for this problem.

Variables:

1. Let $x$ be the horizontal distance between the lake and the point where the pipelines meet.
2. Each manufacturing plant has pipelines that connect to the meeting point, and the total distance from the lake to this meeting point is the variable we need to optimize.

Geometry:

• The plants are 12 km apart.
• The lake is 10 km away horizontally from the plants.
• The meeting point between the pipelines lies somewhere on the 12 km horizontal line connecting the two plants.

Let’s compute the distances of the pipelines using the Pythagorean theorem.

Total Length of the Pipelines:

If the meeting point is $x$ km from the lake along the horizontal axis:

1. Distance from the upper plant to the meeting point is: $\sqrt{x^2 + (12/2)^2} = \sqrt{x^2 + 6^2} = \sqrt{x^2 + 36}.$

2. Distance from the lower plant to the meeting point is the same (by symmetry): $\sqrt{x^2 + 36}.$

3. Total distance from the meeting point to the lake is $10 - x$.

The total length $L$ of the pipelines is: $L(x) = 2\sqrt{x^2 + 36} + (10 - x).$

Optimization:

To find the minimum, we compute $\frac{dL}{dx} = 0$.

1. Differentiate $L(x)$: $L'(x) = 2 \cdot \frac{1}{2} \cdot \frac{2x}{\sqrt{x^2 + 36}} - 1.$ Simplify: $L'(x) = \frac{x}{\sqrt{x^2 + 36}} - 1.$

2. Solve $L'(x) = 0$: $\frac{x}{\sqrt{x^2 + 36}} = 1.$

   Square both sides: $x^2 = x^2 + 36.$

import sympy as sp

Define the variable and the function

x = sp.symbols('x') L = 2 * sp.sqrt(x**2 + 36) + (10 - x)

Differentiate the function

dL_dx = sp.diff(L, x)

Solve the derivative equal to zero

critical_points = sp.solve(dL_dx, x) critical_points[2sqrt(3)][2sqrt(3)]successThe critical point is $x = 2\sqrt{3} \approx 3.464 \, \text{km}$. This represents the distance from the lake to the meeting point of the pipelines that minimizes the total pipeline length.

Verifying Minimum:

To confirm this is a minimum, we check the second derivative:

$L''(x) = \text{second derivative of } L(x).$

Let’s calculate and verify.# Compute the second derivative second_derivative = sp.diff(dL_dx, x)

Evaluate the second derivative at the critical point

second_derivative_at_critical = second_derivative.subs(x, 2 * sp.sqrt(3)) second_derivative_at_criticalsqrt(3)/8sqrt(3)/8successThe second derivative at $x = 2\sqrt{3}$ is positive ($\frac{\sqrt{3}}{8} > 0$), confirming that the total length of the pipeline is minimized at this point.

Conclusion:

The optimal meeting point of the pipelines is approximately 3.464 km from the lake. This minimizes the total length of the pipelines.

Would you like me to explain any step in more detail or visualize the problem? Here are some related questions to consider:

1. How can we compute the exact length of the pipeline after finding the optimal point?
2. Why does minimizing this function involve finding the critical points and checking the second derivative?
3. Can this solution be generalized to other configurations with different plant distances or lake positions?
4. What is the significance of symmetry in this problem, and how does it simplify the calculations?
5. How does the result change if the distance from the lake to the pipeline changes?

Tip: Always confirm a minimum by using the second derivative or evaluating the function at nearby points.