Problem 10 - Calculus and its applications

Question

Find the extremum of \(f(x, y)\) subject to the given constraint, and state whether it is a maximum or a minimum. $$ f(x, y, z)=x^{2}+y^{2}+z^{2} ; x+y+z=2 $$

Step-by-Step Solution

Answer

Minimum value of \( f(x, y, z) \) is \( \frac{4}{3} \) at \( x = y = z = \frac{2}{3} \).

Step 1: Understand the Problem

We are asked to find the extremum (either maximum or minimum) of the function \( f(x, y, z) = x^2 + y^2 + z^2 \) given the constraint \( x + y + z = 2 \). This is a typical problem in optimization involving constraints.

Step 2: Form the Lagrangian

To solve this problem using the method of Lagrange multipliers, we form the Lagrangian function: \[ \mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda(2 - x - y - z) \] where \( \lambda \) is the Lagrange multiplier.

Step 3: Calculate Partial Derivatives

Find the partial derivatives of the Lagrangian with respect to \(x\), \(y\), \(z\), and \(\lambda\), and set them to zero. - \( \frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial z} = 2z - \lambda = 0 \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 2 - x - y - z = 0 \)

Step 4: Solve the Equations

From the partial derivatives, we have the equations:1. \( 2x - \lambda = 0 \)2. \( 2y - \lambda = 0 \)3. \( 2z - \lambda = 0 \)4. \( x + y + z = 2 \).From equations (1), (2), and (3), we conclude that \( x = y = z = \frac{\lambda}{2} \). Substitute in equation (4):\[ 3\frac{\lambda}{2} = 2 \Rightarrow \lambda = \frac{4}{3} \Rightarrow x = y = z = \frac{2}{3} \]

Step 5: Verify and Classify the Extrema

Substituting \( x = y = z = \frac{2}{3} \) back into the function, we find:\[ f\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right) = 3\left(\frac{2}{3}\right)^2 = 3 \times \frac{4}{9} = \frac{4}{3} \]Since the function \( f(x, y, z) = x^2 + y^2 + z^2 \) is convex (second-order derivatives are always positive), the solution \( (\frac{2}{3}, \frac{2}{3}, \frac{2}{3}) \) represents a minimum.

Key Concepts

Optimization with Constraints

When dealing with an optimization problem, constraints often need to be considered. In this exercise, the function we need to optimize is subject to a constraint, specifically, the equation \( x + y + z = 2 \). This is an example of a constraint acting as a restriction on the values that the variables \( x, y, \) and \( z \) can take.
To solve such problems, the method of Lagrange multipliers is commonly used. This involves adding a special variable, called a Lagrange multiplier, to account for the constraint while optimizing the function. We incorporate this constraint into the Lagrangian function:

• \( \mathcal{L}(x, y, z, \lambda) = x^2 + y^2 + z^2 + \lambda(2 - x - y - z) \)

This Lagrangian effectively combines the original function and the constraint, enabling us to find the solution where both are satisfied simultaneously. By differentiating this Lagrangian with respect to the original variables and setting these derivatives to zero, we ensure that any solution we find respects the constraint.
Hence, using Lagrange multipliers, we not only solve for the extremum of functions but do so while respecting the given constraints.

Extremum of a Function

Finding the extremum of a multivariable function involves determining the points at which the function reaches its maximum or minimum values. In this scenario, the function is \( f(x, y, z) = x^2 + y^2 + z^2 \). By using partial derivatives, we find the critical points where changes to any individual variable hold steady.
The critical point for our function, given the constraint \( x + y + z = 2 \), was determined using Lagrange multipliers. The solution, \( x = y = z = \frac{2}{3} \), turned out to be the point at which the extremum occurs. Checking these values in the function gives \( f\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}\right) = \frac{4}{3} \).
Extrema can either be local (limited to a small region around the point) or global (the highest or lowest value across the entire domain of the function). The specificity of a constraint directly influences the type of extremum we identify. Understanding this is crucial when narrowing down solutions, especially when practical applicability is a key goal.

Convex Functions

A convex function is one where any line segment joining two points on the graph of the function remains above the graph itself. Mathematically, this property is indicated by the function having non-negative second-order derivatives.
For the function \( f(x, y, z) = x^2 + y^2 + z^2 \), each second-order partial derivative is positive, affirming its convexity. This implies that the function has a single minimum point, as opposed to having multiple minima, typical of non-convex functions. The relevance of this characteristic is that it guarantees that any extremum found is a global minimum, simplifying the process of identifying whether solutions are maximum or minimum points.
Convex functions are particularly appealing in optimization because their straightforward behavior makes finding solutions more predictable. When optimizing such functions with constraints, as in our exercise, the calculated extremum (where values were found to be \( \frac{2}{3} \) for each variable) is confirmed to be a minimum through the function's convex nature.