/*! 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 55 Find the maximum value of \(s=x ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the maximum value of \(s=x y+y z+x z\) where \(x+y+z=6\).

Short Answer

Expert verified
The maximum value of \(s = xy + yz + zx\) is 12 when \(x = y = z = 2\).

Step by step solution

01

Identify and Express the Objective Function

The objective function that needs to be maximized is given as \(s = xy + yz + zx\). This function combines variables \(x\), \(y\), and \(z\).
02

Set Up the Constraint Equation

The constraint provided in the problem is \(x + y + z = 6\). This is an equation of a plane which binds the values of \(x\), \(y\), and \(z\).
03

Use Lagrange Multipliers

To find the maximum value of \(s\) under the constraint \(x + y + z = 6\), we use the method of Lagrange multipliers. Form the Lagrangian as \( \mathcal{L}(x, y, z, \lambda) = xy + yz + zx + \lambda (6 - x - y - z)\).
04

Find Partial Derivatives

Compute the partial derivatives of \( \mathcal{L} \) with respect to \(x\), \(y\), \(z\), and \(\lambda\) and set them to zero: \( \frac{\partial \mathcal{L}}{\partial x} = y + z - \lambda = 0 \), \( \frac{\partial \mathcal{L}}{\partial y} = x + z - \lambda = 0 \), \( \frac{\partial \mathcal{L}}{\partial z} = x + y - \lambda = 0 \), and \( \frac{\partial \mathcal{L}}{\partial \lambda} = 6 - x - y - z = 0 \).
05

Solve the System of Equations

Solve the system of equations obtained from the partial derivatives: \(y + z = x + z = x + y\) and \(x + y + z = 6\). These equations imply that \(x = y = z\), leading to the substitution \(x = y = z = 2\) after solving \(3x = 6\).
06

Calculate the Maximum Value

Substitute \(x = 2, y = 2, z = 2\) back into the objective function: \(s = xy + yz + zx = 2 \cdot 2 + 2 \cdot 2 + 2 \cdot 2 = 12\). Thus, the maximum value of the function is \(12\).

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Key Concepts

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

Constrained Optimization
Constrained optimization is a technique used to find the maximum or minimum of a function subject to a constraint. When working with real-world problems, constraints often arise naturally. For instance, you might want to maximize profit subject to a budget or resource limitations. In mathematical terms, constrained optimization involves both an objective function and a constraint equation.
In our specific problem, we aim to maximize the function \( s = xy + yz + zx \) while ensuring that \( x + y + z = 6 \). This constraint bounds the possible values of \( x, y, \) and \( z \) such that their sum must be exactly 6. By using methods like Lagrange multipliers, we can transform this constrained problem into one where we can find and evaluate possible extrema.
Objective Function
An objective function is the mathematical expression that you want to optimize, either maximizing or minimizing its value. It is the heart of any optimization problem. In this problem, our objective function is \( s = xy + yz + zx \). This function combines the products of variables \( x, y, \) and \( z \).
The goal of optimizing this function under the given constraint \( x + y + z = 6 \) is to find the specific values of \( x, y, \) and \( z \) that yield the highest value of \( s \). In practical scenarios, the objective function represents what you want to achieve or measure, whether it's profit, cost, distance, etc. Understanding the objective function clearly guides us in applying optimization techniques effectively.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus used to understand how a function changes as each variable changes, while other variables are held constant. They are essential in multivariable calculus, especially when dealing with problems involving multiple variables.
When applying the method of Lagrange multipliers, we take the partial derivative of the Lagrangian with respect to each variable. In our case, the Lagrangian \( \mathcal{L}(x, y, z, \lambda) = xy + yz + zx + \lambda(6 - x - y - z) \) results in partial derivatives like \( \frac{\partial \mathcal{L}}{\partial x} = y + z - \lambda \). Setting these derivatives equal to zero helps us find critical points where the maximum or minimum of the function can occur. It reveals how changes in each variable impact the overall objective function.
System of Equations
A system of equations consists of multiple equations that are solved simultaneously. When using Lagrange multipliers, once the partial derivatives are set to zero, we end up with a system of equations. Solving this system helps us find the values for variables that optimize the function under the given constraints.
In our optimization task, the system \( y + z - \lambda = 0, x + z - \lambda = 0, x + y - \lambda = 0, \) and \( x + y + z = 6 \) arises from the partial derivatives and constraint. These equations imply that each pair of variables added together equals \( \lambda \). Eventually solving \( x = y = z = 2 \) ensures that all conditions are satisfied, and when substituted back they confirm the maximum value of the objective function.

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Most popular questions from this chapter

Find the limits by rewriting the fractions first. $$\lim _{(x, y) \rightarrow(2,0) \atop 2 x-y \neq 4} \frac{\sqrt{2 x-y}-2}{2 x-y-4}$$

Find the maximum value of \(s=x y+y z+x z \quad\) where \(x+y+z=6\).

When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}},\quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right),\quad(3)$$ with all sums running from \(k=1\) to \(k=n\). Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\), 3\. handle data analytically. We demonstrated these ideas with a variety of applications in Section 1.4.

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Function: \(f(x, y)=x y\) Curves: i) The line \(x=2 t, \quad y=t+1\) ii) The line segment \(x=2 t, \quad y=t+1, \quad-1 \leq t \leq 0\) iii) The line segment \(x=2 t, \quad y=t+1, \quad 0 \leq t \leq 1\)

Find and sketch the level curves \(f(x, y)=c\) on the same set of coordinate axes for the given values of \(c .\) We refer to these level curves as a contour map. $$f(x, y)=x+y-1, \quad c=-3,-2,-1,0,1,2,3$$

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