Problem 39 - Calculus and its applications

Question

Find the indicated maximum or minimum values of \(f(x, y)\) subject to the given constraint. Maximum: \(f(x, y, z)=x^{2} y^{2} z^{2} ; x^{2}+y^{2}+z^{2}=2\)

Step-by-Step Solution

Answer

The maximum value is \(\frac{8}{27}\).

Step 1: Set Up Lagrange Multipliers

To find the maximum or minimum values subject to a constraint, use the method of Lagrange multipliers. Define the Lagrangian function:\[ \mathcal{L}(x, y, z, \lambda) = x^2 y^2 z^2 + \lambda (2 - x^2 - y^2 - z^2) \]

Step 2: Find Partial Derivatives

Calculate the partial derivatives of \(\mathcal{L}\) with respect to \(x\), \(y\), \(z\), and \(\lambda\), and set them equal to zero:1. \(\frac{\partial \mathcal{L}}{\partial x} = 2x y^2 z^2 - 2\lambda x = 0\) 2. \(\frac{\partial \mathcal{L}}{\partial y} = 2y x^2 z^2 - 2\lambda y = 0\) 3. \(\frac{\partial \mathcal{L}}{\partial z} = 2z x^2 y^2 - 2\lambda z = 0\) 4. \(\frac{\partial \mathcal{L}}{\partial \lambda} = 2 - x^2 - y^2 - z^2 = 0\)

Step 3: Simplify Equations

Factor out terms and simplify each equation:1. \(2x y^2 z^2 = 2\lambda x \rightarrow y^2 z^2 = \lambda\) if \(x eq 0\)2. \(2y x^2 z^2 = 2\lambda y \rightarrow x^2 z^2 = \lambda\) if \(y eq 0\)3. \(2z x^2 y^2 = 2\lambda z \rightarrow x^2 y^2 = \lambda\) if \(z eq 0\)These imply \(y^2 z^2 = x^2 z^2 = x^2 y^2 = \lambda\). Replace any variable equalities back for confirmation.

Step 4: Solve System of Equations

From solving \(y^2 z^2 = x^2 y^2 = x^2 z^2 = \lambda\), note symmetry among variables implies equal solutions such as \(x = y = z\). Also check \(x^2 + y^2 + z^2 = 2\) with symmetry suggests solutions like \((x, y, z) = \left(\pm \sqrt{\frac{2}{3}}, \pm \sqrt{\frac{2}{3}}, \pm \sqrt{\frac{2}{3}}\right)\).

Step 5: Plug Into Objective Function

Substitute \(x = y = z = \sqrt{\frac{2}{3}}\) or its symmetric variations back into the function \[f(x, y, z) = x^2 y^2 z^2 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}\].

Step 6: Confirm Solution

Double-check all steps and ensure the resultant values are plausible.For any variants where signs change, ensure the function still returns the maximum same value \[f(x, y, z) = \frac{8}{27}\].

Key Concepts

Extrema in Multivariable Calculus

In multivariable calculus, finding extrema involves determining the maximum or minimum values that a function can achieve. Unlike single-variable calculus, multivariable calculus requires navigating through functions of two or more variables, adding complexity to the process. Instead of finding points on a curve, you're discovering points on a surface or even higher-dimensional space.
The task can involve both unconstrained and constrained optimization. In this exercise, we need to consider constraints on the variables, which often leads to using specialized methods like Lagrange multipliers. These techniques allow us to systematically explore the function's behavior under specific limitations, making it possible to find the highest or lowest points that satisfy the given condition.
Understanding multivariable extrema is essential in fields like physics and engineering, where systems often depend on multiple inter-connected parameters. Here, ensuring values meet specific criteria involves analyzing their impact simultaneously in different dimensions.

Constrained Optimization

Constrained optimization focuses on maximizing or minimizing a function subject to specific conditions. In the example given, we seek the maximum of the function \[ f(x, y, z) = x^2 y^2 z^2 \], subject to the constraint that \[ x^2 + y^2 + z^2 = 2 \].
When a constraint is present, solutions must satisfy both the function and the condition simultaneously. This is where the powerful method of Lagrange multipliers comes into play. By introducing a new variable, \( \lambda \), we can convert the problem into solving for critical points of the Lagrangian equation. This equation combines both the function and the constraint, providing a method to locate the maximum or minimum values considering the restriction.

• The Lagrangian function, \( \mathcal{L}(x, y, z, \lambda) \), merges these two aspects into a new entity, thus creating an insightful perspective for solving constrained optimization problems.
• Using the Lagrange multipliers, you derive and analyze partial derivatives to form a system of equations whose solutions provide the required extrema.
• This method works efficiently even with complex multivariable functions, giving a clear path to resolving problems that feature constraints.

Mastering constrained optimization expands the toolbox for dealing with real-world problems in economics, science, and engineering, where conditions are often not freely adjustable.

Partial Derivatives

Partial derivatives are a key concept when dealing with functions of multiple variables. They represent how a function changes as each variable individually changes, while all other variables are held constant. In our exercise, we compute these derivatives to help find the maximum value of the function with respect to given constraints.
When using Lagrange multipliers, you calculate the partial derivatives of both the main function and the constraint, combined in the Lagrangian function, with respect to every involved variable. Here, we calculated:

• \( \frac{\partial \mathcal{L}}{\partial x} \)
• \( \frac{\partial \mathcal{L}}{\partial y} \)
• \( \frac{\partial \mathcal{L}}{\partial z} \)
• \( \frac{\partial \mathcal{L}}{\partial \lambda} \)

Setting these derivatives to zero gives us system equations necessary to find critical points, translating into potential extrema solutions. This makes partial derivatives a fundamental building block in optimization efforts, allowing insights into how changes in one variable can influence the entire system. Through partial derivatives, the behavior of multivariable functions becomes analyzable step by step, leading toward efficient solutions when correctly harnessed.