Problem 14 - College Algebra and Calculus: An Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x, y\), and \(z\) are positive. Maximize \(f(x, y, z)=x y z\) Constraint: \(x+y+z-6=0\)

Step-by-Step Solution

Answer

The maximum of the function \(f(x, y, z) = x y z\), subject to the constraint \(x + y + z = 6\), is at the point (2,2,2).

Step 1: Formulate the Lagrange System

The method of Lagrange multipliers is used to find extrema of a function subject to a constraint. First, set up the Lagrange function: \(L(x, y, z, \lambda) = f(x, y, z) - \lambda (g(x, y, z) - c)\). In this case, it is \(L(x, y, z, \lambda) = x y z - \lambda (x + y + z - 6)\).

Step 2: Differentiate the Lagrange function

The next step involves taking partial derivatives of the Lagrange function with respect to each variable and setting each derivative equal to zero. This yields the following system of equations: \(\partial L / \partial x = yz - \lambda = 0\), \(\partial L / \partial y = xz - \lambda = 0\), \(\partial L / \partial z = xy - \lambda = 0\), \(\partial L / \partial \lambda = x + y + z - 6 = 0\).

Step 3: Solve the System of Equations

Since the first three equations give \(\lambda = yz = xz = xy\), it is clear that \(x = y = z\). Plugging that into the fourth equation \(x + y + z - 6 = 0\) gives \(3x = 6\) and, thus, \(x = y = z = 2\).

Step 4: Verify the solution

Due to the nature of the problem, there are no edges to check for boundary maxima and minima. Hence, the solution obtained in the previous step, (2,2,2), is indeed the maximum.

Key Concepts

Understanding Extremum

An extremum is a point where a function reaches its maximum or minimum value. In optimization problems, finding the extremum involves identifying where a function doesn't increase or decrease, thus hitting either a peak or a trough.

In the context of the exercise, we aim to find the maximum value of the function \( f(x, y, z) = xyz \) under the constraint that \( x + y + z = 6 \). This type of problem usually involves finding critical points, and for multivariable functions, this task can be handled using Lagrange multipliers.

• Maximum: The highest point or value a function can reach in the defined domain.
• Minimum: The lowest point or value a function can reach in the defined domain.
• Critical Points: Points where the derivative (rate of change) of a function is zero or undefined. These are potential candidates for extrema.

Understanding whether a point is a maximum or a minimum requires examining the behavior of the function around the points of interest.

Constraint Optimization

Constraint optimization refers to finding the maximum or minimum of a function while adhering to specific constraints. When using Lagrange multipliers, constraints simplify the optimization by transforming the problem into one that involves solving a system of equations.

In the given exercise, we are maximizing \( xyz \) with the constraint \( x + y + z - 6 = 0 \). This process leads us to leverage the Lagrange function:

\[ L(x, y, z, \lambda) = xyz - \lambda (x + y + z - 6) \]
By introducing a multiplier, \( \lambda \), we convert the constrained problem into one where traditional optimization techniques can be applied.

• Lagrange Multiplier (\( \lambda \)): A factor introduced to convert a constrained problem into an unconstrained one.
• Lagrange Function: The expression \( L(x, y, z, \lambda) \) combines the objective function and the constraint using the multiplier.

Once the Lagrange function is established, the next step is calculating its partial derivatives.

Delving into Partial Derivatives

Partial derivatives are a way to measure how a function changes with respect to one variable while keeping others constant. They are crucial in finding extrema in multivariable functions.

The exercise entails solving the Lagrange function by taking partial derivatives with respect to each variable and the multiplier \( \lambda \) and setting these derivatives to zero. This provides a system of equations that, when solved simultaneously, give the extremum of the function under the constraint.

• \( \frac{\partial L}{\partial x} = yz - \lambda \)
• \( \frac{\partial L}{\partial y} = xz - \lambda \)
• \( \frac{\partial L}{\partial z} = xy - \lambda \)
• \( \frac{\partial L}{\partial \lambda} = x + y + z - 6 \)

These equations allow us to solve for \( x, y, \) and \( z \). Through this methodology, we discover that \( x = y = z = 2 \), meaning the maximum value of \( xyz \) under the constraint is reached when each variable equals 2.

In summary, partial derivatives play a vital role in identifying critical points and ultimately determining the function's extremum when combined with the constraint equation.