Optimized Height Calculation For Maximum Illumination On A Circular Plot

Introduction

In this article, we delve into a fascinating optimization problem involving calculus. Specifically, we aim to determine the optimal height at which to place a light source above a circular plot to maximize the illumination at the edge of the plot. This problem elegantly combines geometric principles with the power of calculus to arrive at a practical solution. Optimization problems are prevalent in various fields, from engineering and physics to economics and computer science. They involve finding the best possible solution from a set of feasible options, often by maximizing or minimizing a certain quantity. Calculus, with its tools for finding rates of change and extrema, provides a robust framework for tackling such problems. Let's embark on this journey to uncover the secrets of optimal illumination.

Problem Statement

Imagine a circular plot of land with a radius, denoted as r, measuring 30 feet. Our task is to position a light source directly above the center of this circular plot. The crucial question is: at what height should we place the light to achieve the maximum illumination at the edge of the plot? This problem involves understanding how light intensity varies with distance and angle, and then using calculus to find the height that optimizes this intensity. The challenge lies in formulating the illumination as a function of the height and then employing calculus techniques, such as finding derivatives and critical points, to determine the maximum value. This classic optimization problem provides a practical application of calculus principles and highlights the importance of mathematical modeling in real-world scenarios.

Setting up the Mathematical Model

To solve this optimization problem, we must first establish a mathematical model that accurately represents the situation. The illumination (I) at a point on the edge of the circular plot depends on two primary factors: the intensity of the light source (which we can assume to be constant) and the distance (d) between the light source and the point. Additionally, the angle (θ) at which the light strikes the edge also plays a crucial role. The illumination is directly proportional to the cosine of the angle of incidence (cos θ) and inversely proportional to the square of the distance (d²). This relationship is derived from the inverse square law of light intensity. Thus, we can express the illumination (I) as:

I = k * (cos θ) / d²

where k is a constant of proportionality that depends on the intensity of the light source. Our goal is to maximize I by adjusting the height (h) of the light source. To do this, we need to express cos θ and d in terms of h and the radius r, which is known. Using the geometry of the situation, we can relate these variables. The distance d can be found using the Pythagorean theorem: d = √(r² + h²). The cosine of the angle θ can be expressed as cos θ = h / d = h / √(r² + h²). Substituting these expressions into the illumination equation, we get:

I = k * h / (r² + h²)^(3/2)

Now we have the illumination I as a function of the height h, which is the variable we can control. The next step is to use calculus to find the value of h that maximizes I. This involves finding the derivative of I with respect to h, setting it equal to zero, and solving for h. The resulting value of h will be the height that provides the maximum illumination at the edge of the circular plot. This mathematical model is crucial for solving the problem and demonstrates how physics and geometry can be combined to create a framework for optimization.

Calculus to the Rescue: Maximizing Illumination

Now that we have our mathematical model, the next step is to use calculus to find the optimized height that maximizes illumination. To do this, we will find the derivative of the illumination function I with respect to the height h, and then find the critical points by setting the derivative equal to zero. The illumination function is given by:

I(h) = k * h / (r² + h²)^(3/2)

where k is a constant and r is the radius of the circular plot (30 ft). To find the derivative I'(h), we will use the quotient rule. The quotient rule states that if we have a function of the form f(h) = u(h) / v(h), then the derivative f'(h) is given by:

f'(h) = [u'(h)v(h) - u(h)v'(h)] / [v(h)]²

In our case, u(h) = k * h and v(h) = (r² + h²)^(3/2). Thus, u'(h) = k and to find v'(h), we use the chain rule:

v'(h) = (3/2) * (r² + h²)^(1/2) * (2h) = 3h(r² + h²)^(1/2)

Applying the quotient rule, we get:

I'(h) = [k(r² + h²)^(3/2) - k * h * 3h(r² + h²)^(1/2)] / (r² + h²)³

We can simplify this expression by factoring out k(r² + h²)^(1/2) from the numerator:

I'(h) = k(r² + h²)^(1/2) * [(r² + h²) - 3h²] / (r² + h²)³

I'(h) = k(r² - 2h²) / (r² + h²)^(5/2)

To find the critical points, we set I'(h) = 0. Since the denominator cannot be zero, we only need to consider the numerator:

r² - 2h² = 0

Solving for h, we get:

2h² = r²

h² = r² / 2

h = ± r / √2

Since height cannot be negative, we take the positive value:

h = r / √2

To confirm that this value of h gives a maximum, we can use the second derivative test or analyze the sign of I'(h) around this point. In this case, it's clear that I'(h) changes from positive to negative at h = r / √2, indicating a maximum. Therefore, the optimized height for maximum illumination is h = r / √2. This calculus-driven approach allows us to precisely determine the ideal height for the light source, showcasing the power of calculus in solving practical optimization problems.

Finding the Optimal Height

Now that we have derived the general formula for the optimized height, h = r / √2, we can substitute the given value of the radius, r = 30 ft, into the equation to find the specific height for this problem. Plugging in the value, we get:

h = 30 / √2

To simplify this expression, we can rationalize the denominator by multiplying both the numerator and the denominator by √2:

h = (30 * √2) / (√2 * √2)

h = (30 * √2) / 2

h = 15√2

Therefore, the optimized height at which the light should be placed above the center of the circular plot to maximize the illumination at the edge is 15√2 feet. Approximating √2 as 1.414, we get:

h ≈ 15 * 1.414

h ≈ 21.21 feet

So, the light should be placed approximately 21.21 feet above the center of the plot to achieve maximum illumination at the edge. This result is a precise and practical solution derived through the application of calculus. It demonstrates how mathematical optimization can provide concrete answers to real-world problems. The process of finding this optimal height involved several steps, from setting up the mathematical model to using calculus to find critical points, and finally substituting the given values to obtain a numerical solution. This comprehensive approach highlights the power and utility of calculus in solving complex optimization problems.

Conclusion

In this article, we successfully tackled an optimization problem to find the optimized height at which to place a light source above a circular plot to maximize illumination at the edge. We started by establishing a mathematical model that related the illumination to the height, radius, and angle of incidence. This model, derived from the inverse square law of light intensity and geometric considerations, formed the foundation for our analysis. We then employed calculus techniques, specifically finding the derivative of the illumination function and setting it equal to zero, to determine the critical points. This process led us to the general formula for the optimized height: h = r / √2, where r is the radius of the circular plot.

Substituting the given radius of 30 feet, we found the specific optimized height to be 15√2 feet, which is approximately 21.21 feet. This result provides a practical solution to the problem, demonstrating how calculus can be used to solve real-world optimization problems. The steps involved in this solution, from setting up the model to applying calculus and interpreting the results, showcase the power and versatility of mathematical methods.

This problem not only provides a concrete example of optimization but also highlights the importance of mathematical modeling in various fields. By translating a real-world scenario into a mathematical framework, we can use powerful tools like calculus to find optimal solutions. Such optimization problems are prevalent in engineering, physics, economics, and many other disciplines, making the understanding and application of calculus crucial for problem-solving and decision-making. The process we followed in this article serves as a valuable template for tackling similar optimization problems in different contexts, emphasizing the enduring relevance of calculus in practical applications.