Problem 14 - Calculus an Applied Approach

Question

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

Step-by-Step Solution

Answer

To maximize \(f(x, y, z) = xyz\) subject to the constraint \(x + y + z - 6 = 0\), one should choose \(x = y = z = 2\), at which point \(f\) attains its maximum value of 8.

Step 1: Setup

Set the given function and constraint. The function that needs to be maximized is \(f(x, y, z) = xyz\). The constraint is \(g(x, y, z) = x + y + z - 6\). Using Lagrange multipliers, define a new function \(L(x, y, z, λ) = f(x, y, z) - λg(x, y, z)\). This results in \(L(x, y, z, λ) = xyz - λ(x + y + z - 6)\).

Step 2: Compute the Partial Derivatives and Set Them Equal to Zero

Compute the partial derivatives of \(L\) with respect to \(x, y , z, λ\), and set each equal to zero to find critical points. Doing so results in the following set of equations:\(yz - λ = 0\),\(xz - λ = 0\),\(xy - λ = 0\),\(x + y + z - 6 = 0\).

Step 3: Solve the System of Equations

Solve the system of equations derived in the previous step. The first three equations imply that \(yz = xz = xy\), which suggests either \(x = y = z\), or one of \(x, y, z\) is zero. Since the problem states that \(x, y, z\) are positive, it must be the case that \(x = y = z\). Substituting \(x = y = z\) into the constraint equation gives \(3x - 6 = 0\), resulting in \(x = 2\). Therefore, \(x = y = z = 2\).

Step 4: Compute the Value of the Function at the Critical Point

Substitute the critical point into the function \(f\). This gives \(f(2, 2, 2) = 2 * 2 * 2 = 8\).

Key Concepts

Understanding Extremum in Mathematics

In mathematics, an extremum refers to the maximum or minimum value of a function. Extremum is central to understanding how a function behaves under various conditions. When we talk about finding the maximum or minimum, we often use terms like optimization.
For example, if you have a box and you want its volume to be as large as possible, its maximum volume would be the extremum. In our exercise, we deal with finding the maximum product of three numbers under a specific constraint.
This concept often connects to critical points and involves derivatives, as these help us understand where maximum and minimum points might occur on a graph. In optimization problems, especially involving constraints, it's all about finding these special points efficiently.

Partial Derivatives and Their Role

Partial derivatives are like regular derivatives but used for functions with multiple variables. Instead of observing how a function changes as we tweak the entire input, partial derivatives focus on changing one variable at a time.
This notion helps simplify complex problems by breaking them down to more manageable parts. In the original exercise, we calculate the partial derivatives of each variable and the Lagrange multiplier (λ). We derived equations such as:

• \( yz - λ = 0 \)
• \( xz - λ = 0 \)
• \( xy - λ = 0 \)
• \( x + y + z - 6 = 0 \)

Setting these to zero helps identify critical points, which are potential extrema. This strategic approach in calculus is instrumental for solving multidimensional problems where direct methods may not be sufficient. Always remember, derivatives provide the rate of change, guiding us to where the function reaches a peak or trough.

Constrained Optimization Using Lagrange Multipliers

When dealing with optimization problems that are not free to vary but have constraints, we use Lagrange multipliers. This is a tool that combines your goal (like maximizing volume) with the limitations you have (like the sum of sides adding up to a particular number).
Suppose you need to maximize a function while adhering to a constraint. The Lagrange multiplier method allows you to find where these two purposes align. The magic happens by constructing a new function called the Lagrangian, incorporating both the target function and the constraint with a multiplier denoted by λ. In our exercise, the Lagrangian is:

• \( L(x, y, z, λ) = xyz - λ(x + y + z - 6) \)

By taking partial derivatives of this new function and setting them to zero, you get a system of equations, providing the paths to find both the optimal values and the nature of the extrema. Constrained optimization using this method thus offers a powerful way to balance different factors in practical problems.