Problem 15 - Calculus an Applied Approach

Question

Use Lagrange multipliers to find the given extremum. In each case, assume that \(x, y,\) and \(z\) are positive. $$ \begin{array}{l}{\text { Minimize } f(x, y, z)=x^{2}+y^{2}+z^{2}} \\ {\text { Constraint: } x+y+z=1}\end{array} $$

Step-by-Step Solution

Answer

The minimum value of the function \(f(x, y, z) = x^2 + y^2 + z^2\) subject to the constraint \(x + y + z = 1\) is \(\frac{1}{3}\), which is achieved at the point \(\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)\).

Step 1: Setting Up the Lagrange Function

The Lagrange function is given by \(L(x, y, z, \lambda) = f(x, y, z) - \lambda(g(x, y, z)-c)\), where \(f(x, y, z)\) is the funcition to be extremized, \(g(x, y, z)\) is the constraint function and \(c\) is the constant. For the given exercise, the Lagrangian becomes \(L(x, y, z, \lambda) = x^2 + y^2 + z^2 - \lambda(x + y + z - 1)\).

Step 2: Setting up the System of Equations

The extremum of the function given the constraint is found at the points where the gradients of the original function and the constraint are collinear. This corresponds to setting the partial derivatives of the Lagrangian equal to zero. The system of equations becomes: \[\begin{align*}\frac{\partial L}{\partial x} = 2x - \lambda = 0 \\frac{\partial L}{\partial y} = 2y - \lambda = 0 \\frac{\partial L}{\partial z} = 2z - \lambda = 0 \x + y + z = 1\end{align*}\]

Step 3: Solving the System of Equations

Solving the first three equations gives \(x=y=z=\frac{\lambda}{2}\). Substituting into the fourth equation gives \(\frac{3\lambda}{2} = 1\). Therefore, \(\lambda = \frac{2}{3}\), and \(x=y=z=\frac{1}{3}\). The minimum value of the function is then found by substituting back into the original function: \(f(x, y, z) = (\frac{1}{3})^2 + (\frac{1}{3})^2 + (\frac{1}{3})^2 = \frac{1}{3}\).

Key Concepts

Extremum

In mathematical analysis, an extremum is a point at which a function's value is at a maximum or minimum. When it comes to multivariable functions, such as \( f(x, y, z) = x^2 + y^2 + z^2 \), determining the extremum becomes more complex. This is because there may be multiple variables affecting the function's output, and the extremum can be a local minimum or maximum, or even an absolute one, depending on the surrounding values of the function.

In the context of our exercise, we are tasked with finding the minimum value of the function given certain constraints. To do so, we employ the method of Lagrange multipliers, which is a strategy for finding the extreme values of a multivariable function subject to constraints that the function must satisfy. This method turns the problem into one of solving a system of equations derived from partial derivatives, which leads us to the extremum under the given constraints.

Constraint Optimization

The idea of constraint optimization is central to many fields such as economics, engineering, and physics. Constraint optimization is about finding the best solution to a problem that includes some restrictions or limits—referred to as constraints. In mathematics, these constraints can often be represented by equations or inequalities that the variables in the function must satisfy.

For the exercise at hand, the constraint is a simple equation \( x + y + z = 1 \), implying that the sum of the variables must equal 1. This constraint shapes the feasible region—a set of points for which the objective function can be evaluated—hence narrowing down the search for the extremum. Lagrange multipliers provide a systematic approach to dealing with such constraint-based problems, enabling us to include the constraints directly into the optimization process.

Partial Derivatives

The concept of partial derivatives plays a pivotal role in multivariable calculus, particularly in optimization problems like the one presented. A partial derivative represents the rate of change of a function as one variable changes while all other variables are held constant. Symbolically, if \( f \) is a function of \( x \) and \( y \) variables, the partial derivative of \( f \) with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \).

When applying the method of Lagrange multipliers, we use partial derivatives to determine the gradient of the function and the gradient of the constraint. By setting these partial derivatives equal to each other, we create a system of equations. Solving this system allows us to find the values of the variables that will give us the extremum of the function within the limits imposed by the constraint. The process is integral to identifying the points where the objective function and the constraint have the same direction of steepest ascent or descent, which aligns perfectly for optimizing the function given the constraint.