Maximize Product With Constraint Find Two Numbers Summing To 228

In the realm of mathematical optimization, one frequently encounters problems that require maximizing or minimizing a certain quantity, subject to given constraints. These types of problems are not only academically interesting but also have wide-ranging applications in fields such as economics, engineering, and computer science. This article delves into a classic optimization problem: finding two positive numbers that satisfy a linear constraint while maximizing their product. We will explore the problem-solving process, highlighting the use of algebraic techniques and calculus to arrive at the solution. This exploration will help in understanding the core principles of optimization and their practical implications.

The core of our discussion lies in a specific mathematical challenge: to identify two positive numbers that adhere to a particular condition while maximizing their product. The condition is that the sum of twice the first number and three times the second number must equal 228. The objective is to determine the values of these two numbers that yield the largest possible product. This problem exemplifies a constrained optimization scenario, where we seek to optimize (in this case, maximize) a function subject to a specific constraint. This type of problem is fundamental in various fields, including economics, engineering, and computer science, where resources are limited and optimal solutions are sought. Understanding how to approach and solve such problems is crucial for making informed decisions and achieving desired outcomes effectively.

Defining the Variables and Objective Function

To begin, let's denote the two positive numbers as x and y. Our objective is to maximize the product of these numbers, which we can express as a function P = x y. This is the objective function we aim to maximize. The problem comes with a constraint: the sum of twice the first number (x) and three times the second number (y) is 228. Mathematically, this constraint can be written as 2x + 3y = 228. This equation provides a relationship between x and y, limiting the possible pairs of numbers we can consider. The challenge now is to find the specific values of x and y that satisfy this constraint and simultaneously make the product P as large as possible. This setup is a classic example of a constrained optimization problem, where we seek to optimize a function (the product P) under a given condition (the constraint 2x + 3y = 228). Solving such problems often involves using techniques from calculus and algebra to find the optimal solution.

Expressing the Objective Function in Terms of a Single Variable

To tackle the optimization problem, the crucial step is to express the objective function, P = x y, in terms of a single variable. This simplifies the problem and allows us to use single-variable calculus techniques for optimization. We achieve this by utilizing the constraint equation, 2x + 3y = 228, to solve for one variable in terms of the other. Let's solve for y in terms of x: 3y = 228 - 2x, which gives us y = (228 - 2x) / 3. Now, we substitute this expression for y into the product function P: P = x * ((228 - 2x) / 3). This results in P as a function of x alone: P(x) = (228x - 2x²) / 3. This transformation is a key step in solving constrained optimization problems, as it reduces the problem to finding the maximum of a single-variable function. With P expressed as a function of x, we can now employ calculus techniques, such as finding critical points, to determine the value of x that maximizes the product P.

Finding the Critical Points Using Calculus

With the objective function P expressed in terms of a single variable x, we can now employ calculus to find the critical points. Critical points are crucial in optimization as they represent potential locations of maxima or minima. To find these points, we first differentiate P(x) with respect to x. Given P(x) = (228x - 2x²) / 3, the derivative P'(x) is calculated as follows: P'(x) = (228 - 4x) / 3. Critical points occur where the derivative is either zero or undefined. In this case, P'(x) is defined for all x, so we focus on finding where it equals zero. Setting P'(x) = 0, we get (228 - 4x) / 3 = 0, which simplifies to 228 - 4x = 0. Solving for x, we find x = 228 / 4 = 57. This value of x is a critical point of the function P(x). To confirm whether this critical point corresponds to a maximum, minimum, or neither, we can use the first or second derivative test. In the next step, we'll use the second derivative test to determine the nature of this critical point.

Determining the Maximum Value Using the Second Derivative Test

After finding the critical point x = 57, it's essential to determine whether this point corresponds to a maximum, a minimum, or an inflection point of the function P(x). The second derivative test is a powerful tool for this purpose. It involves calculating the second derivative of the function and evaluating it at the critical point. If the second derivative is negative at the critical point, it indicates a local maximum; if it's positive, it indicates a local minimum; and if it's zero, the test is inconclusive. For P(x) = (228x - 2x²) / 3, we found the first derivative P'(x) = (228 - 4x) / 3. Now, we differentiate P'(x) to find the second derivative, P''(x). Differentiating (228 - 4x) / 3 with respect to x gives us P''(x) = -4 / 3. Notice that the second derivative is a constant and is negative. Therefore, P''(57) = -4 / 3, which is less than zero. This result confirms that the critical point x = 57 corresponds to a local maximum of the function P(x). This means that the product P is maximized when x is 57. In the next step, we will find the corresponding value of y and the maximum product.

Finding the Corresponding Value of y and the Maximum Product

Having determined the value of x that maximizes the product, which is x = 57, the next step is to find the corresponding value of y. We use the constraint equation 2x + 3y = 228 to solve for y. Substituting x = 57 into the equation, we get 2(57) + 3y = 228, which simplifies to 114 + 3y = 228. Subtracting 114 from both sides gives 3y = 114, and dividing by 3 yields y = 38. So, the value of y that corresponds to the maximum product is 38. Now that we have both x and y, we can calculate the maximum product P. The product P is given by P = x y, so P = 57 * 38 = 2166. Therefore, the maximum product of the two numbers, subject to the given constraint, is 2166. This result completes the solution to the optimization problem, providing the values of x and y that maximize the product while adhering to the constraint.

In conclusion, by employing algebraic manipulation and calculus techniques, we successfully found the two positive numbers that satisfy the given constraint and maximize their product. The two numbers are x = 57 and y = 38, yielding a maximum product of 2166. This problem exemplifies the application of optimization principles in mathematics. Such principles are not only crucial in academic settings but also in real-world scenarios where efficient resource allocation and maximization of outcomes are essential. The methodology used here, involving expressing the objective function in terms of a single variable, finding critical points, and using the second derivative test, is a standard approach in solving constrained optimization problems. Understanding and applying these techniques is vital for anyone involved in decision-making processes across various disciplines, from economics and finance to engineering and computer science. This exploration highlights the power of mathematical tools in solving practical problems and optimizing outcomes in diverse fields.