Problem 34 - Algebra A Combined Function

Question

Solve. Find two numbers whose sum is 11 and whose product is as large as possible.

Step-by-Step Solution

Answer

The numbers are 5 and 6.

Step 1: Define Variables and Equations

Let the two numbers be \(x\) and \(y\). We know from the problem statement that the sum of these two numbers is 11, which gives us the equation \(x + y = 11\).

Step 2: Express One Variable in Terms of the Other

Using the equation \(x + y = 11\), we can express \(y\) in terms of \(x\): \(y = 11 - x\). This substitution will help us express the product of the two numbers in terms of a single variable.

Step 3: Write the Equation for the Product

The product of the two numbers is given by the equation \(P = x \cdot y\). Substituting the expression for \(y\) from Step 2, we get \(P = x(11 - x) = 11x - x^2\).

Step 4: Determine the Critical Points Using Derivatives

To maximize the product, we need to find the critical points of the function \(P(x) = 11x - x^2\) by calculating the derivative. The derivative is \(P'(x) = 11 - 2x\). Set the derivative equal to zero to find the critical points: \(11 - 2x = 0\).

Step 5: Solve for the Maximum Point

Solving \(11 - 2x = 0\) for \(x\), we get \(x = 5.5\). Since \(x\) and \(y\) must be integers, we can check the integers near 5.5, which are 5 and 6.

Step 6: Evaluate the Product for Integer Values

Substitute \(x = 5\) and \(y = 11 - 5 = 6\): the product is \(P = 5 \cdot 6 = 30\). For \(x = 6\) and \(y = 11 - 6 = 5\), the product is also \(P = 6 \cdot 5 = 30\). Both pairs (5, 6) and (6, 5) yield the same maximum product.

Key Concepts

Sum and Product of Numbers

The sum and product of numbers are fundamental concepts often explored in algebra and calculus. In many problems, you are given a constraint such as the sum of two numbers which you need to adhere to, while manipulating another quantity, like the product.
The sum is simply the result of adding two or more numbers together. For instance, if you have two numbers, 5 and 6, their sum is 11. This simple equation can be useful in various mathematical problems as a constraint to solve for unknowns.
The product refers to the result of multiplying these numbers. In our example, the product of 5 and 6 is 30. The challenge often lies in maximizing or minimizing this product under given conditions like a fixed sum.
To solve these types of problems, it might be necessary to express one variable in terms of the other using the sum condition. Then you form an expression for the product in terms of a single variable. The resulting expression becomes a function of one variable, which allows for further exploration, such as finding maximum or minimum values.

Critical Points in Calculus

Critical points are vital in understanding where a function reaches its maximum or minimum values. These points occur where the derivative of a function equals zero or does not exist.

• Critical points can reveal where a function changes direction, indicating potential maximum, minimum, or saddle points.
• For a function of one variable, you find the critical points by taking the derivative, setting it equal to zero, and solving for the variable.

For quadratic functions, such as the product function from the exercise, taking the derivative of the quadratic expression helps identify critical points. In our exercise, the product expression is: \[ P(x) = 11x - x^2 \] The derivative is: \[ P'(x) = 11 - 2x \] Setting the derivative equal to zero results in the critical point: \[ 11 - 2x = 0 \] solves to: \[ x = 5.5 \]
However, since practical solutions often require integer values, verification through substitution of nearby integers might be necessary.
Understanding critical points thus assists in identifying possible maximum and minimum values of your function.

Maximizing Functions

Maximizing functions is a common objective, especially in optimization problems. The goal is often to find the input values at which a function reaches its highest value.
To maximize a function, you generally follow these steps:

• Create a function expressing the quantity to be maximized in terms of one or more variables.
• Use calculus techniques to find critical points, which could potentially be maximum points.
• Evaluate the function at these points and compare the results to determine the maximum value.

In our example, the function: \[ P(x) = 11x - x^2 \] is maximized by finding its critical point, as discussed earlier.
After identifying the critical points, specifically for integer constraints, you substitute back into the original equation to check for maximum product values: \[ P(5) = 30 \] and \[ P(6) = 30 \]
Both values produced the same result, indicating they are indeed maximizing the product within the given constraint. Thus, understanding how to maximize functions effectively is crucial in solving optimization problems.