Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

A powerhouse is located on one bank of a straight river that is \(w\) feet wide. A factory is situated on the opposite bank of the river, \(L\) feet downstream from the point \(A\) directly opposite the powerhouse. What is the most economical path for a cable connecting the powerhouse to the factory if it costs \(a\) dollars per foot to lay the cable under water and \(b\) dollars per foot on land \((a>b)\) ?

Step-by-Step Solution

Answer

Connect underwater directly across, then follow land minimizing costs using x = (b * w) / √(a^2 - b^2).

Step 1: Define the Problem

We need to connect a powerhouse to a factory across a river, minimizing cable-laying costs. The straight-line path across the river and on land is not cost-effective due to varying costs underwater and on land.

Step 2: Determine Variables

Let point P be where the cable emerges on the opposite bank. Define the distance from point A to point P on land as x. Underwater, the cable runs directly across the river, while the rest of the cable runs on land to the factory.

Step 3: Express the Cable Length Under Water

The length of the cable under the river is constant and equal to the width of the river, w.

Step 4: Express the Cable Length on Land

The cable's length on land can be described using the distance between point P (on the same side as the factory) and the factory itself. This is a straight line along the ground and is equal to \(L - x\), where x is the distance from point A to P on the opposite bank.

Step 5: Set up Cost Function

The cost of laying the cable consists of underwater and land portions. Cost for underwater is \(a \times w\) and for land is \(b \times (L-x)\). The total cost is therefore \(a \times w + b \times (L-x)\).

Step 6: Differentiate the Cost Function

To find the minimum cost, we take the derivative of the cost function: \(C = a \times \sqrt{w^2 + x^2} + b \times (L-x)\). Finding the derivative with respect to x gives us: \(C' = \frac{a \times x}{\sqrt{w^2 + x^2}} - b\).

Step 7: Find Critical Points

To find where the cost is minimized, set the derivative to zero: \(\frac{a \times x}{\sqrt{w^2 + x^2}} - b = 0\). Solving this, we get: \(x = \frac{b \times \sqrt{w^2 + x^2}}{a}\).

Step 8: Solve for x

Substitute back to solve: \((a^2 - b^2)x^2 = b^2w^2\), \(x = \frac{b \times w}{\sqrt{a^2 - b^2}}\), This gives the optimal point P along the land route.

Step 9: Analyze Feasibility

Interpret the solution. x should be between 0 and L (0 < x < L) to ensure P is between A and the factory. Otherwise, review constraints or cost structure.

Key Concepts

Calculus in Real World Problems

Calculus is not just a subject confined to textbooks; it plays a crucial role in solving tangible and practical issues we encounter daily. Imagine a simple yet realistic scenario where a powerhouse and a factory need to be connected with a cable across a river. This situation is more than an abstract problem—it's an example of calculus being applied to streamline operations and minimize costs in the real world.
By leveraging the principles of calculus, we can determine the most cost-efficient path for laying the cable. This involves understanding the varying costs of laying cable underwater versus on land, and finding a balance that minimizes the total expense.
To solve such a problem, we utilize mathematical modeling to define the scenario explicitly. With this model, calculus provides the tools necessary to evaluate different options and identify which minimizes cost, demonstrating its power in optimizing real-world systems.

Differential Calculus

Differential calculus focuses on the concept of the derivative, which represents the rate of change of a function. In the context of optimizing cable-laying costs, differential calculus helps us determine how changes in the cable path affect the total cost.
As we analyze the problem, we set up a cost function, which depends on the distances the cable covers underwater and on land. This function is expressed in terms of a variable \( x \), which represents the distance between a reference point and where the cable meets the land.

• The cost function is derived based on the costs per foot of cable underwater and on land, weighted by their respective lengths.
• To find the optimal value of \( x \), we calculate the derivative of this cost function. This derivative tells us how the cost changes as \( x \) changes, and is symbolic of the slopes of these paths.
• By setting the derivative to zero, we find the critical points where the cost does not increase or decrease, indicating potential minima or maxima of cost.

Through these steps, differential calculus becomes integral to pinpointing the best possible solution and understanding how small changes can influence the overall expense.

Cost Minimization Problems

Cost minimization problems revolve around finding the most economical way to achieve a desired outcome, such as connection by a cable in this example. By examining different routes and cost structures, we aim to reduce expenses while fulfilling all requirements.
In the case of the cable-laying task, we first express costs via equations that capture expenses under given conditions:

• Underwater cable costs are represented by \( a \times w \), with \( w \) being the river width.
• The land portion costs \( b \times (L-x) \), where \( x \) is the length of cable on land after crossing the river.

Calculus helps refine this approach by allowing us to differentiate the total cost equation. This differentiation identifies critical points that could potentially minimize expenditure, specifically pointing to locations along the path where the shift in cost is zero. Critically, these points must be evaluated within constraints (between the bank and the factory) to ensure practicality.
Thus, solving such a problem equips students with essential tools for tackling similar challenges in engineering, finance, and various optimization tasks they might encounter in their careers.