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Chapter 11: Problem 8

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y, z)=x^{2} y^{2} z^{2} ; \quad x^{2}+y^{2}+z^{2}=1 $$

Short Answer

Expert verified
Maximum is \( \frac{1}{27} \); minimum is 0.

Step by step solution

01

Define the Function and Constraint

The function we need to optimize is \( f(x, y, z) = x^2 y^2 z^2 \). The constraint is given by \( g(x, y, z) = x^2 + y^2 + z^2 = 1 \).
02

Form the Lagrangian

To apply the method of Lagrange multipliers, we form the Lagrangian function: \( \mathcal{L}(x, y, z, \lambda) = x^2 y^2 z^2 + \lambda(1 - x^2 - y^2 - z^2) \).
03

Compute Partial Derivatives

Calculate the partial derivatives of the Lagrangian:- \( \frac{\partial \mathcal{L}}{\partial x} = 2xy^2z^2 - 2\lambda x \).- \( \frac{\partial \mathcal{L}}{\partial y} = 2x^2yz^2 - 2\lambda y \).- \( \frac{\partial \mathcal{L}}{\partial z} = 2x^2y^2z - 2\lambda z \).- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 1 - x^2 - y^2 - z^2 \).
04

Set Derivatives to Zero

Set each of these partial derivatives to zero to find critical points:1. \( 2xy^2z^2 - 2\lambda x = 0 \).2. \( 2x^2yz^2 - 2\lambda y = 0 \).3. \( 2x^2y^2z - 2\lambda z = 0 \).4. \( 1 - x^2 - y^2 - z^2 = 0 \).
05

Solve the System of Equations

From equations 1, 2, and 3:- If \( x eq 0 \), then \( y^2z^2 = \lambda \). Similarly, if \( y eq 0 \), then \( x^2z^2 = \lambda \), and if \( z eq 0 \), then \( x^2y^2 = \lambda \).- Set these equal to each other: \( y^2z^2 = x^2z^2 = x^2y^2 \), leading to \( x^2 = y^2 = z^2 \).
06

Determine Critical Points

Given \( x^2 = y^2 = z^2 \) and the constraint \( x^2 + y^2 + z^2 = 1 \), solve for the critical points. This implies \( 3x^2 = 1 \), so \( x = \pm \frac{1}{\sqrt{3}} \). Hence, the points are \( (\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}) \).
07

Calculate Function Values at Critical Points

Compute \( f(x, y, z) \) at these points:- \( f\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \).- \( f\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \).
08

Consider Potential Minimum Values

For potential minimum values, examine the edges where one or two coordinates are zero, e.g., \((1, 0, 0)\), yielding \( f(1, 0, 0) = 0 \).
09

Determine Maximum and Minimum Values

Compare calculated values:- The maximum value of \( f(x, y, z) \) is \( \frac{1}{27} \) occurring at critical points from Step 6.- The minimum value is 0 occurring at points like \( (1,0,0) \), \( (0,1,0) \), etc.

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Key Concepts

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

Multivariable Calculus
Multivariable calculus extends the concepts of calculus from one-dimensional functions to functions of several variables. This involves differentiation and integration of functions with multiple inputs, like our function of three variables:
  • Objective Function: This is the function we want to optimize, \( f(x, y, z) = x^2 y^2 z^2 \).
  • Constraints: These are equations that limit the domain of the objective function, such as \( g(x, y, z) = x^2 + y^2 + z^2 = 1 \).
In this context, each variable represents a different dimension of a space, allowing for the exploration of more complex behaviors compared to single-variable calculus. It lays the groundwork for optimization techniques that help find optimal values in various dimensions.
Optimization Problems
Optimization problems are designed to find the best solution under given conditions. In calculus, this often means finding maximum or minimum values of a function.

Objective and Context

In our exercise, we want to determine the maximum and minimum values of \( f(x, y, z) = x^2 y^2 z^2 \) subject to a spherical constraint. Solving this problem involves certain steps:
  • Identify the function to optimize and its constraint.
  • Formulate the problem using auxiliary variables, e.g., Lagrange multipliers.
  • Derive conditions for optima by setting derivatives to zero.
While solving, students should ensure that solutions comprehend the domain imposed by constraints and consider edge cases to accurately identify potential minima or maxima.
Constraint Optimization
Constraint optimization introduces an additional layer to optimization by limiting the solution space with constraints. Lagrange multipliers is a technique used here to handle the constraints effectively.

Lagrange Multipliers

This method constructs a new function, the Lagrangian, from the original function and its constraints:
  • The Lagrangian \( \mathcal{L}(x, y, z, \lambda) = x^2 y^2 z^2 + \lambda(1 - x^2 - y^2 - z^2) \) combines the objective function and the constraint.
  • Partial derivatives with respect to each variable and the multiplier \( \lambda \) are taken and set to zero.
  • Solving these equations simultaneously gives the critical points.
This technique simplifies finding critical points as it incorporates constraints directly into the optimization process. It effectively bridges single-variable problems with real-world multivariable applications.

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Most popular questions from this chapter

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. $$ f(x, y)=x^{2}+y^{2} ; \quad x y=1 $$

Find the extreme values of \(f\) on the region described by the inequality. $$ f(x, y)=e^{-x y}, \quad x^{2}+4 y^{2} \leq 1 $$

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane \(x+2 y+3 z=6\).

(a) Two surfaces are called orthogonal at a point of inter- section if their normal lines are perpendicular at that point. Show that surfaces with equations \(F(x, y, z)=0\) and \(G(x, y, z)=0\) are orthogonal at a point \(P\) where \(\nabla F \neq 0\) and \(\nabla G \neq 0\) if and only if $$F_{x} G_{x}+F_{y} G_{y}+F_{z} G_{z}=0$$ at \(P\) (b) Use part (a) to show that the surfaces $$z^{2}=x^{2}+y^{2}$$ and $$x^{2}+y^{2}+z^{2}=r^{2}$$ are orthogonal at every point of intersection. Can you see why this is true without using calculus?

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y+\frac{1}{x}+\frac{1}{y}$$
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