Problem 66 - Introductory Algebra for College Students

Question

Find two numbers whose sum is 200 and whose product is a maximum.

Step-by-Step Solution

Answer

The numbers that give a maximum product and add up to 200 are both 100.

Step 1: Set The Equations

The exercise gives two constraints, which we convert into mathematical equations. Let's say the two numbers are \(x\) and \(y\). From the problem: \(x + y = 200\) (from the sum) and we need to maximize \(xy\) (which is the product)

Step 2: Express One Variable In Terms Of The Other

We need to turn the maximization problem into a function of a single variable. From the first equation, we can express \(y\) in terms of \(x: y = 200 - x\). Then, we substitute \(y\) value into the product \(P = x(200 - x) = 200x - x^2\) which we will call 'Product Equation'

Step 3: Find the Maximum Of The Function

To find the maximum of the function \(P = 200x - x^2\) we take the derivative and set it equal to zero. The derivative of P is \(P' = 200 - 2x\). Setting it equal to zero we find the value for \(x\) that maximizes the function: \(200 - 2x = 0\) Solving for \(x\) gives \(x = 100\).

Step 4: Find the Corresponding Value of y

Substitute \(x = 100\) into the first equation \(x + y = 200\), we get the corresponding value for \(y\) thus, \(100 + y = 200\). This gives \(y = 100\).

Step 5: Verification

To confirm if we have obtained the maximum product, we can perform a quick check, taking values of \(x\) slightly less than or more than 100 and calculating \(P\). If the values turn out to be lesser than when \(x = 100\), then we can be confident our solution is correct.

Key Concepts

Maximization problems

Maximization problems in algebra typically involve finding the largest or greatest possible value of a function under given constraints. In the context of the original problem, the goal is to maximize the product of two numbers whose sum is a set amount, namely 200 in this case. This is a classic optimization problem that can often be found in various real-world applications such as minimizing costs or maximizing profits in business scenarios.
To tackle a maximization problem, you often start by identifying your equations based on the conditions provided. Here, the key is to determine how one value influences another within those constraints, setting up the necessary equations that guide you toward finding the optimal solution. Typically, these types of problems can be approached in steps:

• Define the variables involved.
• Set up the equations based on the problem statement.
• Transform the problem into a manageable form (often a single-variable function).
• Use calculus to find any critical points which could represent maximum (or minimum) values.
• Verify that the solution meets the original conditions.

For our example, the constraints are the sum of numbers must be 200, and you want to maximize their product. Transforming such problems into a mathematical form is the first crucial step.

Algebraic equations

Algebraic equations are the foundation on which optimization problems are built. They allow us to express relationships between variables in a clear, structured manner. In the exercise, two core equations played a role:

• Sum equation: \(x + y = 200\)
• Product equation: Initially, just \(xy\), later transformed into \(P = x(200 - x) = 200x - x^2\)

By expressing one variable in terms of the other using algebraic manipulation, you simplify the problem into a single-variable equation. This makes it easier to apply calculus tools later on.
Understanding algebraic manipulations is key. You use substitution to replace every instance of one variable, reducing the problem's complexity. Here, substituting \(y = 200 - x\) into the expression for the product helped convert the problem from a two-variable scenario into a one-variable equation. This step is essential not only in algebra but across all problem-solving strategies in mathematics.

Derivatives in algebra

Derivatives are a central concept in calculus, used widely in algebra for solving optimization problems. They allow us to understand the behavior of functions, especially when we want to find maximum or minimum values.
In our exercise, after forming the 'Product Equation' \(P = 200x - x^2\), we take the derivative, \(P' = 200 - 2x\), to find critical points. The derivative tells us how the product changes concerning changes in \(x\). Setting the derivative equal to zero allows us to find the critical point, where the function changes from increasing to decreasing, or vice versa. In simple terms, it's where the maximum or minimum is likely to occur.
Once the critical point, \(x = 100\), is found, it’s important to confirm it's a maximum by testing values around it. Often, small variations around this point will yield smaller product values, confirming it's indeed a peak. Understanding how to derive functions and interpret derivatives is vital in precisely answering maximization and minimization questions in algebra.