Problem 35 - Algebra A Combined Function

Question

Solve. Find two numbers whose difference is 10 and whose product is as small as possible.

Step-by-Step Solution

Answer

The numbers are -5 and 5, with the smallest possible product of -25.

Step 1: Define Variables

Let's define two variables, \( x \) and \( y \), such that these two numbers represent our solution. We'll set \( x \) as the larger number and \( y \) as the smaller number.

Step 2: Set Up Equations

Given that the difference between the numbers is 10, we have the equation \( x - y = 10 \). Our goal is to find the minimum product of these numbers, which can be expressed as \( P = x \times y \).

Step 3: Express One Variable in Terms of the Other

From the equation \( x - y = 10 \), we express \( x \) in terms of \( y \) as \( x = y + 10 \).

Step 4: Substitute and Formulate the Quadratic Equation

Substitute \( x = y + 10 \) into the product equation, giving \( P = y(y + 10) = y^2 + 10y \). This is a quadratic function in terms of \( y \).

Step 5: Determine Critical Points

To find the minimum value of the product \( P = y^2 + 10y \), we take the derivative with respect to \( y \), giving \( \frac{dP}{dy} = 2y + 10 \). Set this to zero to find critical points: \( 2y + 10 = 0 \).

Step 6: Solve for y

Solving \( 2y + 10 = 0 \) gives \( y = -5 \). Since \( x = y + 10 \), substituting \( y = -5 \) gives \( x = 5 \).

Step 7: Verify the Product

Substitute \( y = -5 \) and \( x = 5 \) back into the product formula to verify: \( P = 5 \times (-5) = -25 \). This confirms the minimal product.

Key Concepts

Quadratic Equations

Quadratic equations play a crucial role when solving optimization problems, especially in algebraic contexts. A quadratic equation is typically expressed in the form of \( ax^2 + bx + c = 0 \). In the context of this exercise, we are working with a quadratic function rather than an equation. This function is represented as \( P = y^2 + 10y \) and it helps us explore the relationship between two numbers to find a minimal product.

• The objective of the quadratic function here is to determine how the product varies with changes in \( y \).
• Understanding the structure of the quadratic helps in finding solutions - in this case, to identify the minimum point on a parabola.

Quadratic equations and functions form parabolic shapes, and can possess key points such as vertices, which can represent points of optimization as we encounter here.

Critical Points

Critical points are pivotal when looking for the extrema, either the minimum or maximum values, of a function. For any function, a critical point occurs where its derivative is zero or undefined. This is primarily because, at these points, the slope of the tangent line to the curve is horizontal, indicating a potential peak or trough.

• To find the critical points in our quadratic function \( P = y^2 + 10y \), we calculated its derivative \( \frac{dP}{dy} = 2y + 10 \).
• Setting \( 2y + 10 = 0 \) helped us find the critical value of \( y \).

This technique of finding critical points is tantamount to unlocking the key feature of the function that leads to optimization, particularly in this scenario where we're hunting for a minimum product.

Derivative

Understanding derivatives is essential in the world of calculus and algebraic optimization. A derivative represents the rate at which a function is changing at any given point and is a fundamental tool for finding critical points.

• In our step-by-step solution, the derivative \( \frac{dP}{dy} = 2y + 10 \) indicates how our product \( P \) changes with changes in \( y \).
• This derivative helps us determine where the product is either increasing or decreasing, given \( y \).

When you set a derivative to zero, it tells you where the function's slope is zero, providing a base for identifying potential optima. Mastering derivatives opens the doors to tackling more complex optimization problems effectively.

Minimum Value

Finding a minimum value is a common goal in optimization problems. It refers to the smallest value that a function can attain under given constraints, such as our product \( P = x \times y \) being minimized subject to \( x - y = 10 \).

• The solution involves using the critical point found by setting the derivative to zero, specifically giving us \( y = -5 \).
• Substituting \( y = -5 \) back to compute \( x = y + 10 \), we find \( x = 5 \).

Finally, verifying by calculating \( P = 5 \times (-5) = -25 \), we ensure we've indeed found the minimum product possible under the problem's constraints. Recognizing minimum values in algebra can be significantly aided by visualizing functions and their turning points.