Problem 322 - Calculus Volume 1

Question

Consider the construction of a pen to enclose an area. Two poles are connected by a wire that is also connected to the ground. The first pole is \(20 \mathrm{ft}\) tall and the second pole is \(10 \mathrm{ft}\) tall. There is a distance of \(30 \mathrm{ft}\) between the two poles. Where should the wire be anchored to the ground to minimize the amount of wire needed?

Step-by-Step Solution

Answer

Anchor the wire approximately 16.27 ft from the first pole to minimize wire length.

Step 1: Understanding the Problem

The problem asks us to minimize the total length of a wire connecting two poles, one 20 ft tall and the other 10 ft tall, with a 30 ft distance between them. The wire needs to be anchored to the ground at some point "x" feet from the base of the first pole (20 ft). Our goal is to find this point "x".

Step 2: Expressing the Wire Length (Objective Function)

The wire will be composed of two segments: one from the first pole to the anchor point on the ground, and the other from this anchor point to the second pole. We can use the Pythagorean theorem to express these two wire lengths. The total length of the wire is given by:\[L = \sqrt{x^2 + 20^2} + \sqrt{(30-x)^2 + 10^2}\]

Step 3: Calculating the Derivative of the Objective Function

To minimize the wire length, we need to find the derivative of the objective function with respect to "x" and set it to zero. Take the derivative:\[\frac{dL}{dx} = \frac{x}{\sqrt{x^2 + 400}} - \frac{(30-x)}{\sqrt{(30-x)^2 + 100}}\]

Step 4: Setting the Derivative to Zero and Solving for 'x'

We solve for "x" by setting the derivative to zero:\[\frac{x}{\sqrt{x^2 + 400}} = \frac{(30-x)}{\sqrt{(30-x)^2 + 100}}\]This can be expanded and solved for x. It often involves setting the cross-multiplication equal and solving a quadratic equation.

Step 5: Simplifying and Solving the Equation

Simplifying, we equate:\[x \cdot \sqrt{(30-x)^2 + 100} = (30-x) \cdot \sqrt{x^2 + 400}\]Squaring both sides to remove the square roots and solving gives us a quadratic equation in terms of "x" which can be solved using algebraic methods or a calculator.

Step 6: Finding Critical Points and Determining the Minimum

After solving the quadratic equation, we find the critical points. We should evaluate the length at these critical points and determine which provides the minimum length based on the context of the problem.

Step 7: Verification and Conclusion

Verify the found value by checking if it provides a lesser total wire length compared to endpoints 0 and 30 (as potential extreme points) and ensure it makes practical sense in the problem context.

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand how things change. It is used for finding quantities like distance, area, and volume through processes called differentiation and integration. In this exercise, calculus helps us minimize the length of wire needed. By applying calculus principles, we find the optimal point where the wire should be anchored in order to use the least amount of wire. This involves forming a function that represents wire length and then determining its minimum value.

The steps in a calculus-based optimization problem often include:

• Understanding the problem and what needs to be minimized or maximized
• Setting up an objective function that represents the situation
• Taking the derivative of the function
• Finding critical points by setting the derivative to zero
• Testing these points to determine which leads to the optimal solution

This structured approach is key in solving real-world problems efficiently.

Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate of change or slope of a function. In this optimization problem, we use the derivative to find out how the wire length changes with the position of its anchor on the ground.

To find the optimal wire anchoring point, we first express the length of the wire as a function of "x" and then compute its derivative. The critical step is setting the derivative to zero, which helps find the point that minimizes or maximizes a function.

In practical terms, finding where the derivative equals zero gives us potential solutions where the rate of change is momentarily paused, indicating possible minimum values (which we need for the shortest wire length). Here's a simplified summary:

• Differentiate the objective function
• Set the derivative (rate of change) equal to zero
• Solve for the variable to find critical points

This process is at the heart of solving many optimization problems.

Pythagorean Theorem

The Pythagorean theorem is a crucial tool in this problem for determining wire length segments. The theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

For the wire connected to each pole, we can visualize two triangles formed by the wire, one segment of which runs to the base of each pole. This involves using the theorem:

• For the first wire segment: \( ext{Length} = \sqrt{x^2 + 20^2} \)
• For the second wire segment: \( ext{Length} = \sqrt{(30-x)^2 + 10^2} \)

By applying the Pythagorean theorem, the total wire length is derived as the sum of the lengths of both segments. Accurate usage of this theorem is crucial for setting up the function that calculus is used to optimize.

Quadratic Equation

Quadratic equations frequently arise in optimization problems, especially when solving for critical points. In this exercise, a quadratic equation forms when eliminating square roots from the problem's derivative.

A quadratic equation is generally represented as: \ ax^2 + bx + c = 0 \ , where "a," "b," and "c" are constants. Solving a quadratic can be done using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The process includes:

• Re-arranging the function into a standard quadratic form
• Applying the quadratic formula to find possible values of "x"
• Selecting the solution(s) that meet the problem's real-world constraints

In this context, solving the quadratic gives us potential positions for the wire anchor that minimizes the wire needed between poles. It is an essential step that requires careful computation and understanding of quadratic theory.