/*! 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 9 Minimize \(Q=2 x^{2}+3 y^{2}\), ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 9

Minimize \(Q=2 x^{2}+3 y^{2}\), where \(x+y=5\)

Short Answer

Expert verified
The minimum value of Q is 30 when x = 3 and y = 2.

Step by step solution

01

Define the constraint

Given the constraint equation, express one variable in terms of the other. The constraint is given as: \( x + y = 5 \). Rearrange to find \( y \) in terms of \( x \): \( y = 5 - x \).
02

Substitute constraint into objective function

Substitute the expression for \( y \) from the constraint into the objective function \( Q = 2x^2 + 3y^2 \). Thus, \( Q = 2x^2 + 3(5 - x)^2 \).
03

Simplify the function

Simplify the expression obtained by substituting \( y \): \[ Q = 2x^2 + 3(25 - 10x + x^2) \]. Expanding and combining like terms gives: \[ Q = 2x^2 + 75 - 30x + 3x^2 \] \[ Q = 5x^2 - 30x + 75 \].
04

Find the derivative

To minimize the function, first find the derivative with respect to \( x \): \( \frac{dQ}{dx} = 10x - 30 \).
05

Set the derivative to zero

Set the derivative equal to zero to find the critical points: \( 10x - 30 = 0 \). Solving for \( x \) gives: \( x = 3 \).
06

Calculate corresponding y value

Using the value of \( x \), find the corresponding \( y \) value using the constraint: \( y = 5 - x = 5 - 3 = 2 \).
07

Verify the minimum

Substitute \( x = 3 \) and \( y = 2 \) back into the objective function to verify that it is indeed a minimum: \[ Q = 2(3)^2 + 3(2)^2 \] \[ Q = 18 + 12 = 30 \].

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.

Understanding Calculus in Constrained Optimization
Calculus is a powerful branch of mathematics focused on how things change. It helps us see the relationship between variables and how they fluctuate. In this exercise, we use calculus to find the minimum value of a function. Specifically, we aim to minimize a quadratic function given a linear constraint.
The Minimization Problem
A minimization problem seeks to find the lowest value of a function. Here, our goal is to reduce the function \( Q = 2x^2 + 3y^2 \) to its minimum value.

Let's break it down:
  • We start by expressing one variable in terms of the other using the constraint \( x + y = 5 \).

  • Next, substitute this expression into the original function.

  • After simplification, we achieve a new function in terms of a single variable.

By focusing on a single variable, the problem becomes easier to solve, as it boils down to basic calculus.
Role of Derivatives in Minimization
Derivatives lie at the heart of calculus and minimization problems. They help us determine how a function behaves and identify points where it achieves its lowest or highest values.

In this exercise, we differentiate the function to find its critical points:
  • First, compute the derivative of the simplified function \( Q \) with respect to \( x \).

  • Then, set the derivative equal to zero. This step finds where the slope of the function becomes flat, indicating potential minimum points.

  • Finally, solve for \( x \) and use the constraint to determine \( y \).

This process effectively guides us to the minimum value of the function.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the absolute maximum and minimum values of the function, if they exist, over the indicated interval. When no interval is specified, use the real line \((-\infty, \infty)\). $$ f(x)=2 x^{4}-4 x^{2}+2 $$

Find the limit, if it exists. $$ \lim _{x \rightarrow-\infty} \frac{-6 x^{3}+7 x}{2 x^{2}-3 x-10} $$

The death rate \(r\) per 100,000 males may be modeled as a function of the number of hours per day \(x\) that the men slept. This equation is given by $$ r(x)=104.5 x^{2}-1501.5 x+6016 $$ At how many hours of sleep per day is the death rate minimized?

A pitcher's earnedrun average (the average number of runs given up every 9 innings, or 1 game) is given by $$ E=9 \cdot \frac{n}{i}, $$ where \(n\) is the number of earned runs allowed and is the number of innings pitched. Suppose that we fix the number of earned runs allowed at 4 and let \(i\) vary. We get a function given by $$ E(i)=9 \cdot \frac{4}{i}. $$ a) Complete the following table, rounding to two decimal places. $$ \begin{array}{|l|l|} \hline \begin{array}{l} \text { Innings } \\ \text { Pitched (i) } \end{array} & \begin{array}{l} \text { Earncd-Run } \\ \text { Average (E) } \end{array} \\ \hline 9 & \\ \hline 8 & \\ \hline 7 & \\ \hline 6 & \\ \hline 5 & \\ \hline 4 & \\ \hline 3 & \\ \hline 2 & \\ \hline 1 & \\ \hline \frac{2}{3} \text { (2 outs) } & \\ \hline \frac{1}{3} \text { (l out) } & \\ \hline \end{array} $$ b) Find \(\lim _{i \rightarrow 0^{+}} E(i)\). c) On the basis of parts (a) and (b), determine a pitcher's earned-run average if 4 runs were allowed and there were 0 outs.

Minimize \(Q=x^{3}+2 y^{3}\), where \(x\) and \(y\) are positive numbers, such that \(x+y=1\).
See all solutions

Recommended explanations on Biology Textbooks

Biological Processes

Read Explanation

Cell Communication

Read Explanation

Genetic Information

Read Explanation

Chemistry of Life

Read Explanation

Cellular Energetics

Read Explanation

Reproduction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.