/*! 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 34 Find the absolute maxima and min... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 34

Find the absolute maxima and minima of the functions on the given domains. \(T(x, y)=x^{2}+x y+y^{2}-6 x\) on the rectangular plate \(0 \leq x \leq 5,-3 \leq y \leq 3\)

Short Answer

Expert verified
Absolute maximum: 19; Absolute minimum: -11.

Step by step solution

01

Find Critical Points

First, we need to find the critical points of the function inside the domain. We do this by finding the partial derivatives of the function with respect to both \(x\) and \(y\), and then setting those derivatives equal to zero.\( \frac{\partial T}{\partial x} = 2x + y - 6 \)\( \frac{\partial T}{\partial y} = x + 2y \) Setting these equations to zero, we get:1. \( 2x + y - 6 = 0 \)2. \( x + 2y = 0 \)Solving these simultaneous equations, we find the critical point \((x, y) = (2, -1)\).
02

Evaluate Critical Points

Next, substitute the critical point \((2, -1)\) into the function \(T(x, y)\) to find the value at this point:\( T(2, -1) = (2)^2 + 2(-1) + (-1)^2 - 6(2) = 4 - 2 + 1 - 12 = -9 \).This gives us the function value \( T(2, -1) = -9\).
03

Evaluate Function on the Boundary

Evaluate the function on the boundaries of the rectangular domain. 1. For \(x = 0\), \(0 \leq y \leq 3\), \(T(0, y) = y^2\). Evaluating at the endpoints: - \(T(0, 3) = 9\) - \(T(0, -3) = 9\)2. For \(x = 5\), \(0 \leq y \leq 3\), \(T(5, y) = 25 + 5y + y^2 - 30 = y^2 + 5y - 5\). - \(T(5,3) = 9 + 15 - 5 = 19\) - \(T(5,-3) = 9 - 15 - 5 = -11\)3. For \(y = -3\), \(0 \leq x \leq 5\), \(T(x, -3) = x^2 - 3x + 9\). - \(T(0, -3) = 9\) - \(T(5, -3) = 25 - 15 + 9 - 30 = -11\)4. For \(y = 3\), \(0 \leq x \leq 5\), \(T(x, 3) = x^2 + 3x + 9 - 6x\). - \(T(0, 3) = 9\) - \(T(5, 3) = 19\)
04

Compare all Function Values

Compare all the calculated values of \(T(x,y)\) at the critical point and boundaries:- Critical point: \(T(2, -1) = -9\)- Boundary: \(T(0, 3) = 9\), \(T(0, -3) = 9\), \(T(5, 3) = 19\), \(T(5, -3) = -11\)The absolute maximum is \(19\) and the absolute minimum is \(-11\).

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.

Partial Derivatives
To find the critical points of a function in calculus, partial derivatives are your go-to tools. A partial derivative represents how the function changes as you vary one variable while keeping others constant. For the function given in the exercise, you need to compute the derivative with respect to each variable, which in this case are \(x\) and \(y\).

  • When you compute \( \frac{\partial T}{\partial x} \), you're looking at how the function changes as \(x\) changes, ignoring any change in \(y\).
  • Similarly, \( \frac{\partial T}{\partial y} \) observes the function's change with \(y\) while holding \(x\) constant.
These derivatives help indicate where the function might have peaks (maximums) or valleys (minimums). In this example:\[\frac{\partial T}{\partial x} = 2x + y - 6\]\[\frac{\partial T}{\partial y} = x + 2y\]Setting these to zero helps locate potential places where changes aren't happening, marking potential critical points.
Critical Points
Once you have your partial derivatives, the next step is finding the critical points, which are where these derivatives equal zero. These points are crucial because they can signal peaks, valleys, or saddle points of the function.

To solve for critical points:\[2x + y - 6 = 0\]\[x + 2y = 0\]
  • Simultaneously solving these equations gives \((x, y) = (2, -1)\).
This point is worth investigating further because it might indicate a location or value of interest within the domain where the function behaves differently than elsewhere. Evaluating the function at this point lets us know the exact value the function takes on there.
Boundary Evaluation
Boundary evaluation involves assessing the function's behavior at the edges of the given domain. Since a rectangular domain is specified, these edges are represented by the lines where \(x\) and \(y\) take on their minimum and maximum values. Evaluating a function on all these edges is important because sometimes extremes happen right at the boundary, not just inside.

Let's break it down:
  • For each fixed value (like \(x=0\) or \(y=3\)), substitute into the function and check within the allowed range for the other variable.
  • Perform needed calculations to find \(T(x,y)\) at these endpoints, thus considering all edges of the rectangle.
By calculating these endpoints, like \(T(0, 3) = 9\) or \(T(5, 3) = 19\), you verify whether any boundaries contain maximum or minimum values.
Rectangular Domain
The rectangular domain constrains the space in which you evaluate the function. It's defined by given intervals for \(x\) and \(y\): \(0 \leq x \leq 5\) and \(-3 \leq y \leq 3\). This means when evaluating boundary values, only these segments count.

  • Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi tincidunt orci nec purus sollicitudin cursus.
  • Ut fermentum, velit at maximus molestie, ligula leo vulputate sem, a tristique augue justo sit amet magna.
This domain helps simplify the evaluation only to specific coordinates, making complex problems more manageable by focusing only on parts relevant within these limits. This simplification is key to solving for absolute extrema successfully.

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

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$\begin{array}{l}{f(x, y)=5 x^{6}+18 x^{5}-30 x^{4}+30 x y^{2}-120 x^{3}} \\\ {-4 \leq x \leq 3, \quad-2 \leq y \leq 2}\end{array}$$

Use a CAS to plot the implicitly defined level surfaces in Exercises \(73-76 .\) $$ \sin \left(\frac{x}{2}\right)-(\cos y) \sqrt{x^{2}+z^{2}}=2 $$

Box with vertex on a plane Find the volume of the largest closed rectangular box in the first octant having three faces in the coordinate planes and a vertex on the plane \(x / a+y / b+z / c=1\) where \(a>0, b>0,\) and \(c>0\).

Use a CAS to plot the implicitly defined level surfaces in Exercises \(73-76 .\) $$ 4 \ln \left(x^{2}+y^{2}+z^{2}\right)=1 $$

Find the value of \(\partial z / \partial x\) at the point \((1,1,1)\) if the equation $$x y+z^{3} x-2 y z=0$$ defines \(z\) as a function of the two independent variables \(x\) and \(y\) and the partial derivative exists.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

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.