Question

Find the extreme values, if any, of the following four functions. Check whether they are maxima or minima by the determinantal test. $$z=29-\left(x_{1}^{2}-x_{2}^{2}+x_{3}^{2}\right)$$

Short answer

The function has no strict local extrema at \((0, 0, 0)\).

Step-by-step solution

Define the Function

We are given the function \( z = 29 - (x_1^2 - x_2^2 + x_3^2) \). We need to find the extreme values of this function.

Find the First Partial Derivatives

Calculate the first partial derivatives of \( z \) with respect to \( x_1 \), \( x_2 \), and \( x_3 \): 1. \( \frac{\partial z}{\partial x_1} = -2x_1 \)2. \( \frac{\partial z}{\partial x_2} = 2x_2 \)3. \( \frac{\partial z}{\partial x_3} = -2x_3 \).

Set the First Derivatives to Zero

To find critical points, set the first partial derivatives equal to zero:1. \( -2x_1 = 0 \) implies \( x_1 = 0 \).2. \( 2x_2 = 0 \) implies \( x_2 = 0 \).3. \( -2x_3 = 0 \) implies \( x_3 = 0 \).

Find the Critical Point

The critical point is \((x_1, x_2, x_3) = (0, 0, 0)\).

Find the Second Partial Derivatives

Calculate the second partial derivatives:1. \( \frac{\partial^2 z}{\partial x_1^2} = -2 \)2. \( \frac{\partial^2 z}{\partial x_2^2} = 2 \)3. \( \frac{\partial^2 z}{\partial x_3^2} = -2 \)Mixed partial derivatives are zero.

Form the Hessian Matrix

The Hessian matrix \( H \) at \((0, 0, 0)\) is:\[H = \begin{bmatrix}-2 & 0 & 0 \0 & 2 & 0 \0 & 0 & -2 \\end{bmatrix}\]

Apply Determinantal Test

Compute the determinants of the leading principal minors:1. \( D_1 = |-2| = -2 \)2. \( D_2 = \left|\begin{array}{cc} -2 & 0 \ 0 & 2 \end{array}\right| = (-2)(2) - (0)(0) = -4 \)3. \( D_3 = \det(H) = |-8| = -8 \). Since the determinants alternate in sign, the function has neither a local minimum nor maximum at \((0,0,0)\).

Conclude the Analysis

The determinantal test reveals that the critical point does not correspond to a local extremum (neither maximum nor minimum).

Key concepts

Extreme Values

When we talk about extreme values, we are referring to the highest or lowest points on a function, known as maximum and minimum points. Identifying these points in a multivariable function involves finding where the function reaches its largest or smallest value. These critical points occur where the slope of the function is zero, meaning the function isn't increasing or decreasing.To find the extreme values of a function, we first calculate the first derivative or, in the case of functions with several variables, the first partial derivatives. Then, we set these derivatives equal to zero to find the critical points. These points are significant because they are potential candidates for extreme values where the function might reach a peak or dip.In the given exercise, the function is evaluated at the critical point \(0, 0, 0\), where all first partial derivatives are zero. However, determining whether these critical points are actually extreme values requires further investigation using tests like the determinantal or other criteria.

Hessian Matrix

The Hessian matrix offers a powerful way to understand the curvature of a multivariable function near critical points. It is composed of the second partial derivatives of the function—essentially, it helps us understand how the slopes of the tangent planes change.In our exercise, the Hessian matrix is formed using the second derivatives of the given function, which is evaluated at the critical point. The structure of the Hessian matrix looks like this: \[H = \begin{bmatrix}\frac{\partial^2 z}{\partial x_1^2} & \frac{\partial^2 z}{\partial x_1 \partial x_2} & \frac{\partial^2 z}{\partial x_1 \partial x_3} \\frac{\partial^2 z}{\partial x_2 \partial x_1} & \frac{\partial^2 z}{\partial x_2^2} & \frac{\partial^2 z}{\partial x_2 \partial x_3} \\frac{\partial^2 z}{\partial x_3 \partial x_1} & \frac{\partial^2 z}{\partial x_3 \partial x_2} & \frac{\partial^2 z}{\partial x_3^2}\end{bmatrix}\]In this case, evaluating the Hessian matrix helps in determining the nature of the critical point through the determinantal test. By looking at the sign of the determinants of its leading principal minors, we can decide if a critical point is a local maximum, minimum, or a saddle point.

Partial Derivatives

Partial derivatives are like regular derivatives but for functions with multiple variables. They measure the rate of change of the function with respect to one variable while keeping the others constant. They are essential for finding the slope and curvature of multivariable functions.To identify critical points and check for extreme values, we calculate the first partial derivatives of a function concerning each variable. In our exercise, the first step was computing these partial derivatives:- The derivative with respect to \(x_1\) gives \(\frac{\partial z}{\partial x_1} = -2x_1\).- For \(x_2\), it is \(\frac{\partial z}{\partial x_2} = 2x_2\).- And for \(x_3\), \(\frac{\partial z}{\partial x_3} = -2x_3\).By setting these partial derivatives to zero, we find the critical points—in this example, \(x_1 = 0\), \(x_2 = 0\), and \(x_3 = 0\). Partial derivatives also feed into the Hessian matrix to further evaluate the nature of these critical points.