/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Find two numbers \(x\) and \(y\)... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 38

Find two numbers \(x\) and \(y\) with \(x-y=36\) for which the term \(x y\) is minimized.

Short Answer

Expert verified
The numbers are 18 and -18.

Step by step solution

01

Understanding the Problem

We want to find two numbers, denoted by \(x\) and \(y\), such that the difference \(x-y=36\). We need to minimize the product \(xy\).
02

Express One Variable in Terms of the Other

Given \(x - y = 36\), we can express one variable in terms of the other. Let's express \(x\) in terms of \(y\): \[ x = y + 36 \]
03

Write the Product Function

Now that we have \(x = y + 36\), we can substitute it into the product \(xy\):\[ P = xy = (y + 36)y = y^2 + 36y \]
04

Find the Derivative

To minimize the product, we need to find the critical points by taking the derivative of \(P\). The derivative is:\[ \frac{dP}{dy} = 2y + 36 \]
05

Set the Derivative to Zero

Set the derivative equal to zero to find the critical points:\[ 2y + 36 = 0 \]Solving for \(y\), we get:\[ 2y = -36 \]\[ y = -18 \]
06

Calculate Corresponding \(x\) Value

Using \(y = -18\) in the equation \(x = y + 36\):\[ x = -18 + 36 = 18 \]
07

Verification

The critical point \((x, y) = (18, -18)\) satisfies the original condition \(x-y=36\). To ensure this point is a minimum, notice that a second derivative test would affirm minimizing since it's a parabola opening upwards.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product of Numbers
In many optimization problems, you might come across the concept of finding two numbers whose product needs to be either maximized or minimized. In this instance, we aim to find two numbers, \(x\) and \(y\), such that their difference equals 36 and their product \(xy\) is minimized. Understanding the product of numbers in this context helps us to set up the equations we need to solve the problem efficiently.

When given such a task:
  • Use the condition provided (e.g., \(x-y=36\)) to express one variable in terms of the other.
  • Substitute into the product equation to form a function of a single variable (e.g., \(P = y(y+36)\)).
  • Use calculus techniques to find the values that optimize this product.
Using this step-by-step approach helps guarantee we are considering all necessary conditions set by the problem.
Critical Points
Finding critical points is a key step in many optimization problems. Once we have the product function, we aim to determine where this function reaches its minimum value. Critical points occur where the derivative of the function is zero or undefined. Here's the basic idea:
  • Take the derivative of the product function \(P\) with respect to \(y\).
  • Set this derivative equal to zero to find potential minimum or maximum values.
For instance, in our problem, the derivative of the product \(P = y^2 + 36y\) is \(\frac{dP}{dy} = 2y + 36\). Setting this equal to zero, \(2y + 36 = 0\), gets us \(y = -18\). Critical points are necessary as they give potential solutions where the function behavior changes.
Derivative
Derivatives are foundational in calculus optimization tasks. A derivative represents the rate of change of a function concerning one of its variables. When we want to find the minimum or maximum of a function, the derivative helps us locate the critical points.
  • The first derivative tells us the slope of the function. When it's zero, the function has a flat region that might be a minimum or maximum.
  • By solving \(\frac{dP}{dy} = 2y + 36 = 0\), we find critical points that are candidates for minima or maxima.
The application of derivatives lets us analyze and confirm which points yield the best optimization result for the function equation derived from the original problem conditions.
Parabola
A parabola is a symmetric curve that one often sees in quadratic functions, like the one we encountered in the form of \(P = y^2 + 36y\). In optimization, understanding the shape of this curve is crucial because the nature of the parabola informs us about whether the critical point is a minimum or maximum.
  • For a quadratic function \(ax^2 + bx + c\), if \(a > 0\), the parabola opens upwards, indicating a minimum point.
  • If \(a < 0\), it opens downwards, pointing to a maximum point.
In our problem, since the parabola \(y^2 + 36y\) has a positive coefficient for \(y^2\), it opens upwards. Thus, the critical point we found at \(y = -18\) indeed provides the minimum product, giving us assurance in our solution.

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Most popular questions from this chapter

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