Problem 165 - Advanced Calculus

Question

Find the extrema for the function \(\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}\) subject to the constraint \(x^{2}+2 y^{2}-z^{2}-1=0\)

Step-by-Step Solution

Answer

The extrema for the given function \(f(x, y, z) = x^2 + y^2 + z^2\) subject to the constraint \(g(x, y, z) = x^2 + 2y^2 - z^2 - 1 = 0\) do not exist. We used the method of Lagrange multipliers to create a system of equations with partial derivatives and attempted to solve for \(x, y, z\), and the Lagrange multiplier \(\lambda\). However, the resulting system of equations did not provide any real solutions for the extrema.

Step 1: Set up the Lagrange multiplier system

The method of Lagrange multipliers states that the gradient of the function must be parallel to the gradient of the constraint. The equation for setting up the Lagrange multiplier system is \(\nabla f = \lambda \nabla g\), which is a system of partial derivatives: \[ \begin{cases} \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y} \\ \frac{\partial f}{\partial z} = \lambda \frac{\partial g}{\partial z} \\ g(x, y, z) = 0 \end{cases} \]

Step 2: Compute the partial derivatives

Compute the partial derivatives for both the function \(f\) and the constraint \(g\): \(\frac{\partial f}{\partial x} = 2x\), \(\frac{\partial f}{\partial y} = 2y\), \(\frac{\partial f}{\partial z} = 2z\) \(\frac{\partial g}{\partial x} = 2x\), \(\frac{\partial g}{\partial y} = 4y\), \(\frac{\partial g}{\partial z} = -2z\)

Step 3: Set up the system of equations

Now we plug the partial derivatives into the equations: \[ \begin{cases} 2x = \lambda (2x) \\ 2y = \lambda (4y) \\ 2z = -\lambda (2z) \\ x^2 + 2y^2 - z^2 - 1 = 0 \end{cases} \]

Step 4: Solve the system of equations for x, y, z, and λ

Let's solve the first three equations for \(\lambda\): 1. \(\lambda = \frac{2x}{2x} = 1\) (assuming \(x \neq 0\)) 2. \(\lambda = \frac{2y}{4y} = \frac{1}{2}\) (assuming \(y \neq 0\)) 3. \(\lambda = -\frac{2z}{2z} = -1\) (assuming \(z \neq 0\)) Since \(\lambda\) cannot have different values simultaneously, we can conclude that \(x = 0\) or \(y = 0\). If \(x = 0\), the constraint becomes \(2y^2 - z^2 - 1 = 0\). And if \(y = 0\), the constraint becomes \(x^2 - z^2 - 1 = 0\). Let's solve these two subcases separately:

Step 4a: Solve for x = 0

When \(x = 0\), the constraint is \(2y^2 - z^2 - 1 = 0\). Since \(\lambda = -1\), the second equation is now \(2y = -4y\), which yields \(y = 0\). Thus, the constraint is now left with \(-z^2 - 1 = 0\), which has no real solutions for \(z\).

Step 4b: Solve for y = 0

When \(y = 0\), the constraint is \(x^2 - z^2 - 1 = 0\). Since \(\lambda = \frac{1}{2}\), the first equation is now \(2x = x\), which yields \(x = 0\). Again, we have the constraint \(-z^2 - 1 = 0\), which has no real solutions for \(z\). Thus, there are no extrema for the given function with the given constraint.

Key Concepts

Extrema of a Function

The search for the extrema of a function is a fundamental problem in calculus, and it revolves around finding the points where a function reaches its highest or lowest values, called maxima and minima, respectively. These points are critical for understanding the behavior of functions, especially in physics and engineering where they can represent states of equilibrium, optimal solutions, and more.

When we talk about unconstrained optimization, the extrema are where the gradient (vector of partial derivatives) of a function is zero. However, when the function is subjected to a constraint, as in the exercise example, traditional methods such as setting the gradient to zero won't directly apply. Here's where the powerful method of Lagrange multipliers steps in, which is designed specifically for constrained optimization scenarios.

System of Partial Derivatives

A system of partial derivatives is a set of equations that represent the rate of change of a multivariable function with respect to each variable independently. It's like taking a snapshot of how sensitive a function is to changes in different directions. These derivatives are the components of the gradient.

In the given exercise, computing the partial derivatives of the function \( f \) and the constraint \( g \) provides us with the necessary components to set up a system of equations that must be satisfied for the function to have an extremum under the given constraint. To find potential extrema, we equate the gradient of \( f \) to a multiple (given by a Lagrange multiplier, \( \lambda \) ) of the gradient of \( g \)—a clever way to ensure that while we’re finding an extremum of \( f \) , we're not straying from the constraint \( g(x, y, z) = 0 \).

Constraint Optimization

Constraint optimization problems are those where the solution must satisfy certain conditions or restrictions, known as constraints. It's like solving a puzzle with specific rules—only certain pieces fit. In mathematics, constraints are expressed as equations or inequalities that limit the feasible solutions.

The method of Lagrange multipliers is an elegant strategy to these problems. It introduces an auxiliary multiplier, \( \lambda \) , which acts as a weight adjusting the influence of the constraint on the function we're trying to optimize. By doing so, it transforms the constrained problem into a system of equations where the gradients of the function and the constraint are proportional—the key insight that allows us to solve for the variables and \( \lambda \).

In our exercise, we applied this method to find that if \( x \) or \( y \) is zero, it would leave us with an unsolvable constraint, leading to the conclusion that there are no real extrema under the given conditions—showing that sometimes the solution to a constraint optimization problem is that there's no possible solution within the given constraints.