Question

Industrial costs Dayton Power and Light, Inc. has a power plant on the Miami River where the river is 800 ft wide. To lay a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs \(\$ 180\) per foot across the river and \(\$ 100\) per foot along the land. (a) Suppose that the cable goes from the plant to a point \(Q\) on the opposite side that is \(x\) ft from the point \(P\) directly opposite the plant. Write a function \(C(x)\) that gives the cost of laying the cable in terms of the distance x. (b) Generate a table of values to determine if the least expensive location for point \(Q\) is less than 2000 ft or greater than 2000 \(\mathrm{ft}\) from point \(P .\)

Short answer

The total cost function of laying the cable is: \( C(x) = 180 \cdot \sqrt{800^2 + x^2} + 100 \cdot (2000 - x) \) if \(x \leq 2000\) or \( C(x) = 180 \cdot \sqrt{800^2 + x^2} + 100 \cdot (x - 2000) \) if \(x > 2000\). To determine whether the least expensive location for point \(Q\) is less than or greater than 2000 ft from point \(P\), calculate \(C(x)\) for various values of \(x\).

Step-by-step solution

Create function C(x) across the river

Using the Pythagorean theorem, the length of the cable across the river to point \( Q \) is \( \sqrt{800^2 + x^2} \) feet. Since the cost of laying the cable across the river is $180 per foot, the cost in terms of \(x\), \( C_{river}(x) \), can be calculated as: \( C_{river}(x) = 180 \cdot \sqrt{800^2 + x^2} \).

Create function C(x) along the land

The length of the cable along the land to point \( Q \) is \(2000 - x\) feet (assuming \(x \leq 2000\) feet) or \(x - 2000\) feet (assuming \(x > 2000\) feet). Since the cost of laying the cable along the land is $100 per foot, the cost in terms of \(x\), \( C_{land}(x) \), can be calculated as: \( C_{land}(x) = 100 \cdot (2000 - x) \) if \(x \leq 2000\) or \( C_{land}(x) = 100 \cdot (x - 2000) \) if \(x > 2000\).

Combine functions for the total cost

The total cost of laying the cable, \( C(x) \), is the sum of the cost of laying the cable across the river and along the land, i.e., \( C(x) = C_{river}(x) + C_{land}(x) \).

Generate values and comparison

By substituting several values of \(x\) into the function \(C(x)\), one could get a sense of the cost of laying the cable for several distances. Then, compare whether the cost is less for locations less than or greater than 2000 ft from point P.

Key concepts

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that plays a crucial role in various applications, including calculus. The theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, if we have a right triangle with sides of length 'a' and 'b', and the hypotenuse of length 'c', it is expressed as:
\[ c^2 = a^2 + b^2 \]
In the context of our problem, the cable across the river forms the hypotenuse of a right triangle, with the river width and the distance along the land from point P being the other two sides. Therefore, by using the Pythagorean theorem, we can calculate the length of the cable needed to cross the river and, thus, determine the associated cost. This concept is not only vital in theoretical mathematics but also in practical, real-world problems, such as laying the cable in our exercise.

Total Cost Function

A total cost function combines all the costs associated with a particular undertaking into a single equation. It's a critical tool in economics and business for understanding the complete financial implication of any action. In calculus, a total cost function is a function that takes various inputs, like the quantity of goods produced or, in our exercise, the distance 'x' where the cable will land on the other side of the river, to calculate the overall expense.

In our example, the total cost function accounts for the two types of costs incurred—laying the cable across the river and along the land. By expressing these costs in terms of 'x', we can easily calculate the total expense of the operation for different values of 'x'. This function helps in determining the most cost-effective point Q to lay the cable. Understanding how to formulate and manipulate such functions is essential for optimizing outcomes in real-world scenarios.

Cost Analysis

Cost analysis is a process by which an individual or company reviews the financial implications of each decision. This typically involves examining each portion of the costs involved and determining the optimal path that minimizes expense while achieving the desired outcome. It's an invaluable part of budgeting and financial planning.

In the scenario provided, cost analysis involves determining the different costs for laying cable across the river and along the land at varying distances 'x' from point P. By breaking down the total cost into these constituent parts and then summing them, we carry out a detailed cost analysis. This process aims to find the value of 'x' that results in the lowest possible total cost. Effective cost analysis requires careful consideration of all variables, and in this case, calculus provides the necessary tools for a thorough examination.

Calculus Applications

Calculus, a branch of mathematics focused on change and motion, has numerous applications in various fields, including physics, engineering, economics, and everyday problem-solving. In our specific problem, calculus helps in formulating a cost function, analyzing it, and then finding its minimum—to optimize the cable laying project's cost.

Calculus concepts allow us to understand how the total cost varies as 'x' changes and helps in determining the point at which this change will be at its lowest, indicating the most cost-efficient location for point Q. By creating formulas for the cost and then differentiating and finding critical points, we apply calculus techniques to solve real-world optimization problems efficiently. Such applications showcase the power and practicality of calculus, beyond its abstract theoretical base.