Question

Use Lagrange multipliers to find the given extremum of \(f\) subject to two constraints. In each case, assume that \(x, y\), and \(z\) are nonnegative. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) Constraints: \(x+2 z=6, x+y=12\)

Short answer

To carry out this exercise, one must set up the Lagrangian function and then obtain and evaluate the system of equations derived from the partial derivatives. Once the values of \(x, y, z\) are determined, they need to be plugged into the function to find the minimum value, while adhering to the constraints given.

Step-by-step solution

Formulate the Lagrangian function

In order to solve this problem, we need to introduce Lagrange multipliers \(λ\) and \(μ\). The Lagrangian function \(L\) is therefore given by \(L(x, y, z, λ, μ) = f(x, y, z) − λ(g(x, y, z)−6) − μ(h(x, y, z)−12)\), where here \(f(x, y, z) = x^2 + y^2 + z^2, g(x, y, z) = x + 2z\) and \(h(x, y, z) = x + y\). This gives us the Lagrangian function: \(L = x^2 + y^2 + z^2 - λ(x + 2z - 6) - μ(x + y - 12)\)

Obtain and evaluate partial derivative

Take partial derivatives of \(L\) with respect to \(x, y, z, λ, μ\) and equate them to zero to find stationary points: \(/partial L/\partial x = 2x - λ - μ = 0, /partial L/\partial y = 2y - μ = 0, /partial L/\partial z = 2z - 2λ = 0, /partial L/\partial λ = x + 2z - 6 = 0, /partial L/\partial μ = x + y - 12 = 0\)

Solve the system of equations

The system of five equations can be reduced. From the first three equations, we obtain \(x = (λ + μ)/2, y = μ/2, z = λ. Substituting these into the last two equations: λ + μ/2 - 6 = 0, λ + μ - 12 = 0\. Solving this system will yield the values of λ and μ, which can further be substituted into the above equations to get the values of \(x, y, z\) respectively.

Evaluate the obtained (x,y,z) in given function

Finally, insert these \(x, y, z\) values into the function \(f(x, y, z) = x^{2} + y^{2} + z^{2}\). The result will be the minimum value for the function subject to the given constraints.

Key concepts

Extremum

When working with optimization problems, the concept of an "extremum" becomes crucial. Extremum refers to the maximum or minimum value that a function can achieve. In this exercise, we're particularly interested in finding the minimum value of the function \( f(x, y, z) = x^2 + y^2 + z^2 \).
By applying Lagrange multipliers, we aim to find this minimum under certain restrictions, known as constraints.Finding extremum values under constraints involves more advanced techniques because such problems don't have straightforward solutions. The traditional methods of finding minimum or maximum values require modifying the function itself with added constraints, introducing variables called multipliers. These help us determine where the extremum might be considering everything you are required to take into account.

Partial Derivatives

Partial derivatives play an essential role in the method of Lagrange multipliers. Simply put, a partial derivative of a function measures how the function changes as one particular variable is varied, with all other variables held fixed.
In our problem, partial derivatives help us find the stationary points of the Lagrangian function \( L(x, y, z, \lambda, \mu) \). For example, the partial derivative of \( L \) with respect to \( x \), denoted as \( \frac{\partial L}{\partial x} \), is calculated by differentiating \( L \) while keeping \( y \), \( z \), \( \lambda \), and \( \mu \) constant.

• By setting each partial derivative equation to zero, \( \frac{\partial L}{\partial x} = 0 \), \( \frac{\partial L}{\partial y} = 0 \), and so on, we determine potential solutions where the function does not change, suggesting a possible extremum.
• These equations help us build a system to solve for variables that provide the function's extreme values given the constraints.

System of Equations

When solving optimization problems with constraints using Lagrange multipliers, one often encounters a set of simultaneous equations, commonly referred to as a system of equations.
In this scenario, the partial derivatives from the Lagrangian lead us to a system of equations. These equations connect multiples variables: \( x \), \( y \), \( z \), and the multipliers, \( \lambda \) and \( \mu \).

• The system is built from the conditions \( \frac{\partial L}{\partial x} = 0 \), \( \frac{\partial L}{\partial y} = 0 \), \( \frac{\partial L}{\partial z} = 0 \), and the constraint equations \( x + 2z - 6 = 0 \) and \( x + y - 12 = 0 \).
• Solving this system helps determine the values of these variables where the function achieves its extremum given the constraints. This solution provides the specific values that satisfy all relevant conditions.

Constraints

Constraints are conditions or equations that the solution to an optimization problem must satisfy.
In this exercise, we are given two constraints: \( x + 2z = 6 \) and \( x + y = 12 \). Both are linear equations that bind the values \( x \), \( y \), and \( z \) can take.

• Constraints are crucial as they define the feasible region within which we search for the extremum.
• Using Lagrange multipliers accommodates these constraints in the optimization problem, ensuring that the solution is not just optimal in an unconstrained universe but according to the given limitations.
• The constraints ensure that the solution is both realistic and applicable to real-world scenarios, which commonly involve limits and restrictions.