Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 34
Find the indicated maximum or minimum values of \(f\) subject to the given constraint. Maximum: \(f(x, y, z)=x+y+z ; \quad x^{2}+y^{2}+z^{2}=1\)
Short Answer
Expert verified
The maximum value is \(\sqrt{3}\).
Step by step solution
01
Define the Problem
The objective is to find the maximum value of the function \(f(x, y, z) = x + y + z\) subject to the constraint \(x^2 + y^2 + z^2 = 1\).
02
Introduce the Method of Lagrange Multipliers
Use the method of Lagrange multipliers to incorporate the constraint. Define the Lagrangian function \(\mathcal{L}(x, y, z, \lambda) = f(x, y, z) - \lambda (g(x, y, z) - 1)\), where \(g(x, y, z) = x^2 + y^2 + z^2\).
03
Set Up the Lagrangian Function
Write the Lagrangian as follows: \[\mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - 1)\]
04
Find the Partial Derivatives and Set to Zero
Take the partial derivatives of \(\mathcal{L}\) with respect to each variable and the multiplier \(\lambda\). Set each derivative equal to zero to find the critical points:\[\frac{\partial \mathcal{L}}{\partial x} = 1 - 2\lambda x = 0\]\[\frac{\partial \mathcal{L}}{\partial y} = 1 - 2\lambda y = 0\]\[\frac{\partial \mathcal{L}}{\partial z} = 1 - 2\lambda z = 0\]\[\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0\]
05
Solve the System of Equations
From the equations:\(1 - 2\lambda x = 0\), \(1 - 2\lambda y = 0\), and \(1 - 2\lambda z = 0\):\(x = \frac{1}{2\lambda}\), \(y = \frac{1}{2\lambda}\), \(z = \frac{1}{2\lambda}\). Substitute these into the constraint \(x^2 + y^2 + z^2 = 1\):\[\left(\frac{1}{2\lambda}\right)^2 + \left(\frac{1}{2\lambda}\right)^2 + \left(\frac{1}{2\lambda}\right)^2 = 1\] \[\frac{3}{(2\lambda)^2} = 1\] \[\lambda = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}\]
06
Determine Critical Points
Using \(\lambda = \frac{\sqrt{3}}{2}\) and \(\lambda = -\frac{\sqrt{3}}{2}\):\(x = y = z = \frac{1}{\sqrt{3}}\) for \(\lambda = \frac{\sqrt{3}}{2}\)\, and \(x = y = z = -\frac{1}{\sqrt{3}}\) for \(\lambda = -\frac{\sqrt{3}}{2}\).
07
Evaluate the Function at Critical Points
Calculate \(f(x, y, z)\) at each critical point:\(f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = 3 \cdot \frac{1}{\sqrt{3}} = \sqrt{3}\).\(f\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) = 3 \cdot -\frac{1}{\sqrt{3}} = -\sqrt{3}\).
08
State the Maximum Value
The maximum value of the function \(f(x, y, z) = x + y + z\) subject to the constraint is \(\sqrt{3}\).
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.
Maximum Value
In this exercise, our goal is to find the maximum value of the function \( f(x, y, z) = x + y + z \). The concept of maximum value involves finding the highest point that a function can reach within a certain constraint. The constraint here is given by \( x^2 + y^2 + z^2 = 1 \), representing a sphere with a radius of 1.
To find this maximum value, we utilize the method of Lagrange multipliers, which helps in optimizing a function under given constraints. The steps involve, first, defining the function and constraint and then utilizing partial derivatives to find the critical points.
Finally, we evaluate the function at these critical points to determine the maximum value.
To find this maximum value, we utilize the method of Lagrange multipliers, which helps in optimizing a function under given constraints. The steps involve, first, defining the function and constraint and then utilizing partial derivatives to find the critical points.
Finally, we evaluate the function at these critical points to determine the maximum value.
Constraint Optimization
Constraint optimization is crucial in various fields like economics, engineering, and operations research. It allows us to optimize a function subject to certain limitations or constraints. In this exercise, the constraint is a spherical surface, \( x^2 + y^2 + z^2 = 1 \), upon which we need to find the maximum value of \( f(x, y, z) = x + y + z \).
The method of Lagrange multipliers is a powerful technique for constraint optimization. It introduces an auxiliary variable (the multiplier \( \lambda \)) to incorporate the constraint into the function being optimized. By doing this, we convert the problem into a system of equations derived from the Lagrangian function \( \mathcal{L}(x, y, z, \lambda) \).
The Lagrangian function for this optimization problem is \( \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - 1) \).
Solving the system of equations resulting from setting the partial derivatives of \( \mathcal{L} \) to zero gives us the critical points, which we further analyze to find the maximum value of the function.
The method of Lagrange multipliers is a powerful technique for constraint optimization. It introduces an auxiliary variable (the multiplier \( \lambda \)) to incorporate the constraint into the function being optimized. By doing this, we convert the problem into a system of equations derived from the Lagrangian function \( \mathcal{L}(x, y, z, \lambda) \).
The Lagrangian function for this optimization problem is \( \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - 1) \).
Solving the system of equations resulting from setting the partial derivatives of \( \mathcal{L} \) to zero gives us the critical points, which we further analyze to find the maximum value of the function.
Partial Derivatives
Partial derivatives are essential tools when working with functions of several variables. They help us understand how a function changes as each variable changes, while keeping the others constant.
In this exercise, we use partial derivatives to find the critical points of the Lagrangian function \( \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - 1) \).
We take the partial derivatives with respect to each variable \( x \), \( y \), \( z \), and the Lagrange multiplier \( \lambda \):
- \( \frac{\partial \mathcal{L}}{\partial x} = 1 - 2\lambda x = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial y} = 1 - 2\lambda y = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial z} = 1 - 2\lambda z = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0 \)
By solving these equations, we find that \( x = y = z = \frac{1}{2\lambda} \) and we substitute these back into the constraint equation to determine the value of \( \lambda \).
This process guides us to the critical points, which we evaluate to find the function's maximum and minimum values.
In this exercise, we use partial derivatives to find the critical points of the Lagrangian function \( \mathcal{L}(x, y, z, \lambda) = x + y + z - \lambda (x^2 + y^2 + z^2 - 1) \).
We take the partial derivatives with respect to each variable \( x \), \( y \), \( z \), and the Lagrange multiplier \( \lambda \):
- \( \frac{\partial \mathcal{L}}{\partial x} = 1 - 2\lambda x = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial y} = 1 - 2\lambda y = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial z} = 1 - 2\lambda z = 0 \)
- \( \frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 + z^2 - 1 = 0 \)
By solving these equations, we find that \( x = y = z = \frac{1}{2\lambda} \) and we substitute these back into the constraint equation to determine the value of \( \lambda \).
This process guides us to the critical points, which we evaluate to find the function's maximum and minimum values.
