Problem 359 - Calculus Volume 3

Question

Use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints. $$ f(x, y, z)=x y z, x^{2}+2 y^{2}+3 z^{2}=6 $$

Step-by-Step Solution

Answer

Max: \( f = \sqrt{3} \), Min: \( f = -\sqrt{3} \) at \((\sqrt{3}, 1, 1)\) and \((-\sqrt{3}, -1, -1)\).

Step 1: Define the Problem

We need to find the extremum of the function \( f(x, y, z) = x y z \) subject to the constraint \( x^2 + 2y^2 + 3z^2 = 6 \). This is a problem of finding extrema with constraints.

Step 2: Write the Lagrangian

The Lagrangian function \( \mathcal{L} \) is given by combining the objective function and the constraint with a multiplier \( \lambda \): \[ \mathcal{L}(x, y, z, \lambda) = x y z + \lambda (6 - x^2 - 2y^2 - 3z^2) \]

Step 3: Compute the Gradient

Find the gradients \( abla f \) and \( abla g \). Here, \( abla f = (yz, xz, xy) \) and \( abla g = (2x, 4y, 6z) \). For getting extremums, \( abla \mathcal{L} = (abla f - \lambda abla g) = 0 \).

Step 4: Set Up Equations

Equate partial derivatives of \( \mathcal{L} \) to zero:1. \( yz - 2\lambda x = 0 \) 2. \( xz - 4\lambda y = 0 \)3. \( xy - 6\lambda z = 0 \)4. The constraint \( x^2 + 2y^2 + 3z^2 = 6 \).

Step 5: Solve the System of Equations

From equation 1, \( \lambda = \frac{yz}{2x} \).From equation 2, \( \lambda = \frac{xz}{4y} \).From equation 3, \( \lambda = \frac{xy}{6z} \).Equate these expressions for \( \lambda \) and solve with the constraint to find potential \((x, y, z)\).

Step 6: Find Critical Points

Solving the equations, we get potential solutions such as \((\pm\sqrt{3}, \pm 1, \pm 1)\). Substitute back into the constraint to verify they satisfy \( x^2 + 2y^2 + 3z^2 = 6 \).

Step 7: Evaluate Function at Critical Points

Compute \( f(x, y, z) = x y z \) at each critical point obtained. For \((\sqrt{3}, 1, 1)\) and \((-\sqrt{3}, -1, -1)\), \( f(x, y, z) = \sqrt{3} \) and \( f(x, y, z) = -\sqrt{3} \) respectively.

Key Concepts

Extremum Finding

Finding the extremum of a multivariable function involves identifying the points where the function reaches its maximum or minimum values. This is crucial in optimization as it allows us to determine the most efficient or optimal outcomes for different scenarios. In some cases, given a function like \( f(x, y, z) = x y z \), we want to find these points not in an open space but subject to certain conditions, also known as constraints.
This process typically involves taking derivatives and setting them to zero to find critical points, which are potential maximums or minimums. In our context, using the Lagrange multipliers, it helps us efficiently tackle these constraints while searching for extrema. It's like looking for the highest or lowest point on a restricted landscape. Understanding the significance of extremum in calculations lets us solve real-world problems with precision.

• The extremum could be either a maximum or minimum of a function.
• Critical points are places where the gradient (or derivative) is zero or undefined.
• These points can indicate peaks, valleys, or flat points in the context of the function.

Solving for extrema is a fundamental exercise in mathematics and real-world applications, especially when constraints are at play.

Constrained Optimization

Constrained optimization is about finding the optimum (maximum or minimum) value of a function within a specified range or under specified conditions. This exercise is vital when resources or variables in your function are limited or bounded by equations, just like the constraint \( x^2 + 2y^2 + 3z^2 = 6 \) from our example.
Lagrange multipliers help us efficiently handle these constraints by introducing a new variable to merge the constraint into the optimization problem. Let's imagine we try to maximize or minimize a function, akin to an employee looking to maximize productivity given time and material limits. This tool simplifies problems by transforming them into systems we can solve using standard techniques, like equation solving, which involves fundamentals of algebra and calculus.

• Constraints are equations or inequalities that limit the domain of the solution.
• Lagrange multipliers help incorporate these constraints directly into the function we need to optimize.
• This method works by making the gradient of the objective function align with the gradient of the constraint, multiplied by the Lagrange multiplier.

This encapsulation of the constraint within the function through a multiplier offers a powerful method, especially in complex optimization scenarios across various fields, from economics to engineering.

Gradient Equations

Gradient equations are central to understanding optimization problems, especially those involving more than one variable. The gradient of a function tells us the direction of steepest ascent. For the function \( f(x, y, z) = x y z \), its gradient is \((yz, xz, xy)\), showing how the function changes with respect to \(x\), \(y\), and \(z\).
In constrained optimization, gradients guide the alignment between the function and constraint. Using our Lagrangian function, \( \mathcal{L}(x, y, z, \lambda) = xyz + \lambda (6 - x^2 - 2y^2 - 3z^2) \), we extract gradients \( abla f \) and \( abla g \) to form equations that dictate where these should be parallel; \( abla f - \lambda abla g = 0 \). This condition ensures that at the extremum, the path defined by the constraint follows the function's shape, akin to ensuring a cart follows the contour of a hill as best as it can.

• The gradient of a function is a vector that provides the direction of maximum increase of the function.
• In solving \( abla \mathcal{L} = 0 \), the Lagrangian gradient is set to zero to find stationary points.
• Finding solutions involves setting partial derivatives to zero and solving, leading to possible extremum candidates.

Understanding and applying gradient equations is pivotal in solving optimization problems effectively, helping locate points that maximize or minimize a given function under constraints.