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To solve this optimization problem, let's define some variables and work through the cost function that we need to minimize.

### Problem Setup

1. Let $ x $ represent the distance from point $ C $ to point $ P $, along the wall.
2. Given that $ A $ is 19 meters from $ C $, the remaining distance from $ P $ to $ A $ is $ 19 - x $.
3. The vertical distance from $ P $ to $ M $ is 6 meters.

### Cost Calculation

The cost of the cable has two parts:

1. **From $ C $ to $ P $** along the wall, at $8 per meter. This cost is:
   \[
   \text{Cost from } C \text{ to } P = 8 \times x
   \]

2. **From $ P $ to $ M $**, which is the direct (straight-line) distance, at $10 per meter. Using the Pythagorean theorem, the distance from $ P $ to $ M $ is:
   \[
   \text{Distance from } P \text{ to } M = \sqrt{(19 - x)^2 + 6^2}
   \]
   So the cost from $ P $ to $ M $ is:
   \[
   \text{Cost from } P \text{ to } M = 10 \times \sqrt{(19 - x)^2 + 6^2}
   \]

### Total Cost Function

The total cost $ C(x) $ as a function of $ x $ is:
\[
C(x) = 8x + 10 \sqrt{(19 - x)^2 + 6^2}
\]

### Minimizing the Cost

To find the value of $ x $ that minimizes $ C(x) $, we need to take the derivative of $ C(x) $ with respect to $ x $, set it to zero, and solve for $ x $.

1. **Differentiate $ C(x) $ with respect to $ x $:**
   \[
   \frac{dC}{dx} = 8 + 10 \cdot \frac{1}{2} \cdot \frac{2(19 - x)(-1)}{\sqrt{(19 - x)^2 + 6^2}}
   \]

2. **Set $ \frac{dC}{dx} = 0 $ and solve for $ x $.**

Let's calculate it step-by-step.The critical point that minimizes the cost occurs when the distance from $ C $ to $ P $ is 11 meters.

### Calculating the Minimum Cost
Now, let's substitute $ x = 11 $ back into the total cost function to find the minimum cost.The minimum cost of installing the cable is $188 when the distance from $ C $ to $ P $ is 11 meters.

Would you like a more detailed explanation of any step or have further questions?

### Related Questions

1. How would the solution change if the cost per meter from $ C $ to $ P $ was different?
2. What if the point $ M $ was located at a different vertical distance from the wall?
3. How can we confirm that this critical point indeed minimizes the cost?
4. What would happen to the solution if the cost per meter for $ P $ to $ M $ increased?
5. How is the Pythagorean theorem used in real-world optimization problems like this one?

#### Tip
In optimization problems, always double-check the units and whether each part of the cost function is correctly scaled for a real-world scenario.

A cable runs along the wall from C to P at a cost of $8 per meter, and straight from P to M at a cost of $10 per meter. If M is 6 meters from the nearest point A on the wall where P lies, and A is 19 meters from C, find the distance from C to P such that the cost of installing the cable is minimized and find this cost.

Learn how to minimize the cost of installing a cable that runs along a wall from point C to P, then directly to point M. This problem involves calculating the optimal distance from C to P using optimization techniques, including differentiation and the Pythagorean theorem. Suitable for advanced calculus students.

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