Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 11: Problem 35
Solve. Find two numbers whose difference is 10 and whose product is as small as possible.
Short Answer
Expert verified
The numbers are -5 and 5, with the smallest possible product of -25.
Step by step solution
01
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.
02
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 \( P = x \times y \).
03
Express One Variable in Terms of the Other
From the equation \( x - y = 10 \), we express \( x \) in terms of \( y \) as \( x = y + 10 \).
04
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 \( y \).
05
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: \( 2y + 10 = 0 \).
06
Solve for y
Solving \( 2y + 10 = 0 \) gives \( y = -5 \). Since \( x = y + 10 \), substituting \( y = -5 \) gives \( x = 5 \).
07
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.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \( y \).
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 \( y \).
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 \( x = 5 \).
